ACM模板
用于简化模运算的定理
欧拉定理 a与m互质，a^phi(m)%m==1-->m为质数 a^(m-1)%m==1
威尔逊定理 p为素数 (p-1)!%p==p-1==-1
费马定理 p为素数，a和p互质，a^(p-1)%p==1
费马小定理 p为素数，a^p%p==a-->(a^p/a)%p==a/a-->a^(p-1)%p==1
a和m不互质时若m较小可以直接找a^x%m=1,即直接for循环从1开始找x
A^x % m = A^(x%phi(m)+phi(m)) % m (x >= phi(m))
用于化简(a^(b^(c^...))),先算最上面的次方,然后递归回去
换底公式 log(a)(b)=log(c)(b)/log(c)(a)，当结果超过浮点数范围时用log算,负数没有对数
#pragma comment(linker, "/STACK:102400000,102400000")//手动开大栈区
多次模规律 x=x*x%p;在一定次数后x会不再改变，即x*x%p==x
用scanf(“%[^ ]”,s);时记得初始化s[0]=0;否则输入有空行时会WA,且输入空行时返回值为0
a开n次根:x=pow(a,1.0/n)
处理误差时同时判x*x==a || (x+1)*(x+1)==a (间隔为2x+1,所以基本能处理误差)
树上某个点的子树的所有信息为dfs时进入这个点时已经记录的信息和离开这个点时已经记录的信息的差集
//multiset中 s.erase(2)会一次删除所有的 2，若要只删一个应用 s.erase(s.find(2))--易错
多个样例输入 T较大且满足从小到大递推时可以将询问离线出来处理
abs(x)在没有 using namespace std;的情况下超过 long long 会出错
gcd计数相关:莫比乌斯反演,容斥原理,欧拉函数
递归超过 1e5层左右就会爆栈 RE,将递归改成手写栈
1 C(n,k+1)=C(n,k)*(n-k)/(k+1) 2 3 //O(n)递推 4 5 C(n,k)=n!/(k!*(n-k!))=n*(n-1)*...*(n-k+1){k个相乘}/k! (k<=n) 6 7 C(n,k)=C(n,n-k) 8 9 C(n,k)=C(n-1,k)+C(n-1,k-1) 10 11 {杨辉等式，杨辉三角形的对称性} 12 13 C(n,0)+C(n,1)+...+C(n,n-1)+C(n,n)=2^n {二项式定理} 14 15 C(n,0)+C(n+1,1)+C(n+2,2)+.+C(n+r,r)=C(n+r+1,r)//把 C(n,0)看出 C(n+1,0)不断往后合并 16 17 可重复组合（隔板法） 18 19 从 n个元素中可重复地选 k个，n个盒子放 k个相同的球，选法为 20 21 C(n+k-1,n-1)=C(n+k-1,k) 22 23 有重复元素排列 24 25 n个元素中有 n1个相同，n2个相同。。。 26 27 n!/(n1!*n2!*...*nk!) 28 29 圆排列 30 31 从 n个不同元素中选 k个不分首位围成一个圆圈，方案数为 32 33 A(n,k)/k 如果 k==n，则有 n!/n=(n-1)!种 34 35 条件排列 36 37 将 1,2,3,4,5五个数进行排列，要求 4排在 2前面 38 39 方法 1：先排 135,3!种,再插 2,4，先插 4，有四种插法，对于每种插法，2有 4,3,2,1种插法，所以种数=3!*10=60 40 41 方法 2：不考虑 2,4的全排列，共有 5！种，2和 4的位置先后分布情况各半，所以种数=5!/2=60 42 43 错位排列 44 45 {a1,a2,...,an}，ai不放在第 i个位置的种数 46 47 Dn=n!*(1-1/1!+1/2!-1/3!+1/4!-...!(-1)^n/n!) 48 49 f(n)=(n-1)*(f(n-1)+f(n-2)) 递推式
1 一、球相同，盒子相同，且盒子不能空 2 3 结论 ｎ个相同的球放入ｍ个相同的盒子（n≥m），不能有空盒时的放法种数等于ｎ分解为ｍ个数的和的种数. 4 5 二、球相同，盒子相同，且盒子可以空 6 7 结论 ｎ个相同的球放入ｍ个相同的盒子（n≥m），可以有空盒时的放法种数等于将ｎ分解为ｍ个、(ｍ－1)个、 8 9 (ｍ－2)个、…、2 个、1 个数的和的所有种数之和. 10 11 三、球相同，盒子不同，且盒子不能空 12 13 结论 ｎ个相同的球放入ｍ个不同的盒子中（n≥m），不能有空盒的放法数 C(n-1,m-1). 14 15 四、球相同，盒子不同，且盒子可以空 16 17 结论 ｎ个相同的球放入ｍ个不同的盒子中（n≥m），可以有空盒的放法数 C(n+m-1,m-1). 18 19 五、球不同，盒子相同，且盒子不能空 20 21 . 22 23 递推式:S(n,m)=S(n-1,m-1)+m*S(n-1,m) 24 25 结论 ｎ个不同的球放入ｍ个相同的盒子中（n≥m），不能有空盒的放法种数等于ｎ个不同的球分成ｍ堆的种 26 27 数. 28 29 六、球不同，盒子相同，且盒子可以空 30 31 结论 ｎ个不同的球放入ｍ个相同的盒子中（n≥m），可以有空盒的放法种数等于将ｎ个不同的球分成ｍ堆、(ｍ 32 33 －1)堆、(ｍ－2)堆、…、2 堆、1 堆的所有种数之和. 34 35 七、球不同，盒子不同，且盒子不能空 36 37 结论 ｎ个不同的球放入ｍ个不同的盒子中，不能有空盒的放法种数等于ｎ个不同的球分成ｍ堆的种数乘以ｍ！. 38 39 八、球不同，盒子不同，且盒子可以空 40 41 结论 ｎ个不同的球放入ｍ个不同的盒子中（n≥m），可以有空盒的放法种数等于ｍ^ｎ种.
将 n个不同的球和 m个不同球按原来顺序”穿插”起来,即将 n个球看成一样的,m个也是一样的,然后进行排列,有
(n+m)!/(n!m!)中插法,即将 n个球按原来顺序放入(m+1)个盒子
1 前几项:1 1 2 5 14 42 132 429 1430 4862 16796 58786 2080127429002674440969484535357670129644790 2 3 1.递归公式 4 5 h(n) = h(n-1)*(4*n-2) / (n+1); 6 7 2.递归公式 8 9 f(n)=f(0)*f(n-1)+f(1)*f(n-2)+...+f(n-2)*f(1)+f(n-1)*f(0){累加 f(i)*f(n-i-1),i从 0到 n-1} 10 11 3.组合公式 1 12 13 f(n)=C(2*n,n)/(n+1) 14 15 4.组合公式 2 16 17 f(n)=C(2*n,n)-C(2*n,n-1) 18 19 用于 20 21 1 二叉数形态的计数 22 23 2 火车进站后的出站顺序种数,乘法加括号的运算顺序种数 24 25 3 将凸 n边形用 n-3条不相交的对角线分成 n-2个互相没有重叠的三角形的分法 26 27 4 28 29 n个 A和 n个 B排成一排，在任何位置 B的个数不超过 A的个数的排列种数
1 假设 x为奇数，y为偶数，则 z为奇数，2z与 2x的最大公因数为 2，2z和 2x可分别写作 2 3 2z = (z + x) + (z - x) 4 5 2x = (z + x) - (z - x) 6 7 那么跟据最大公因数性质，z + x和 z - x的最大公因数也为 2，又因为： 8 9 (z + x)(z - x) = y^2，两边同除以 4得： 10 11 ((z + x) / 2)((z - x) / 2) = (y / 2)^2 12 13 故可令： 14 15 z + x = 2m^2, z - x = 2n^2 16 17 其中 z = m + n, x = m - n（m与 n互质） 18 19 则有： 20 21 y2 = z^2 - x^2 = (2m)^2 (2n)^2 = 4m^2 n^2 22 23 即 y = 2mn。 24 25 综上所述，可得到下式： 26 27 x = m^2 - n^2, y = 2mn, z = m^2 + n^2. (m, n为任意自然数)
#ifdef ONLINE_JUDGE
#else
freopen("in","r",stdin);
freopen("out","w",stdout);
#endif
1 struct node 2 { 3 int d,i; 4 }d[110]; 5 struct cmp 6 { 7 bool operator ()(const node &a, const node &b) 8 { 9 return a.d<b.d || a.d==b.d && a.i>b.i; 10 } 11 }; 12 priority_queue<node, vector<node>, cmp>q; 13 priority_queue<int, vector<int>, less<int> >q; 14 priority_queue<int, vector<int>, greater<int> >q;
1 template <class T> 2 bool scanff(T &ret) 3 { //Faster Input:short,int,long long,double,float 4 char c; int sgn; T bit=0.1; 5 if(c=getchar(),c==EOF) return 0; 6 while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar(); 7 sgn=(c=='-')?-1:1; 8 ret=(c=='-')?0:(c-'0'); 9 while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0'); 10 if(c==' '||c==' ') { ret*=sgn; return 1; } 11 while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10; 12 ret*=sgn; 13 return 1; 14 }
1 template <class T> 2 void printff(T ret) 3 { //Faster Output:short,int,long long 4 char s[22]; 5 if(ret<0) { printf("-");ret=-ret; } 6 int len=0; 7 while(ret) { s[++len]=ret%10+'0';ret/=10; } 8 if(!len) s[++len]='0'; 9 while(len) printf("%c",s[len--]); 10 }
1 void phi_table(int *phi,int *f,int *p,int &pn) 2 { 3 phi[1]=1;pn=0; 4 int k; 5 for(int i=2;i<N;i++) 6 { 7 if(!f[i]) { p[pn++]=f[i]=i;phi[i]=i-1; } 8 for(int j=0;j<pn&&(k=p[j]*i)<N;j++) 9 { 10 f[k]=p[j]; 11 if(f[i]==p[j]) { phi[k]=phi[i]*p[j]; break; } 12 else phi[k]=phi[i]*(p[j]-1); 13 } 14 } 15 //for(int i=2;i<=N;i++) phi[i]+=phi[i-1];//累加 16 }
1 void fa(int n,int &dn,int*d) 2 { 3 dn=0; 4 int m=sqrt(n)+0.5; 5 for(int i=1;i<=m;i++) if(n%i==0) { d[dn++]=i;d[dn++]=n/i; } 6 if(m*m==n) dn--; 7 sort(d,d+dn); 8 }
1 void fa(int n,int &pn,int *p) 2 { 3 pn=0; 4 int m=sqrt(n)+0.5; 5 for(int i=2;i<=m;i++) if(n%i==0) 6 { 7 p[pn]=1; 8 while(n%i==0) 9 { 10 p[pn]*=i; 11 n/=i; 12 } 13 pn++; 14 } 15 if(n!=1) p[pn++]=n; 16 }
1 void Prin() //O(nlogn) 2 { 3 memset(prin,0,sizeof(prin)); 4 for(int i=2;i<N;i++) if(!prin[i]) 5 for(int j=i;j<N;j+=i) 6 { 7 int t=j; 8 while(t%i==0) 9 { 10 prin[j]++; 11 t/=i; 12 } 13 } 14 }
1 template<class T> 2 T gcd(T a,T b) 3 { 4 while(b) 5 { 6 T r=b; 7 b=a%b; 8 a=r; 9 } 10 return a; 11 }
1 LL slove(LL a,LL b) 2 { 3 LL g = gcd(b/a,a); 4 while(g!=1) 5 { 6 a = a/g; 7 g = gcd(b/a, a); 8 } 9 return b/a; 10 }
1 LL quick_mod(LL t,LL n) 2 { 3 LL ans=1; 4 while(n) 5 { 6 if(n&1) ans=ans*t%mod; 7 t=t*t%mod; 8 n>>=1; 9 } 10 return ans; 11 }
//欧拉函数为 phi(n)=小于 n且与 n互质的数的个数,这些数的和为 n*phi(n)/2
证明: 设 共有 m 个与 n 互质的数, m 为 n 的欧拉函数, m=φ(n)；
设 ans=∑i=0,m−1 ai…(1) ,gcd(ai,n)=1&& ai<n;
可知 gcd(ai,n−ai)=1;
则 ans=∑i=0,m−1 (n−ai)…(2) , gcd(ai,n−ai)=1 && n−ai<n;
(1) + (2) 得 2∗ans=n∗m, 即 ans=n∗φ(n)/2;
欧拉函数引理
1.p为素数 phi(p)=p-1
2.p为素数 phi(p^a)=(p-1)*p^(a-1)=p^a*(1-1/p)
3.a,b互质 phi(a*b)=phi(a)*phi(b)
n=p1^a1*p2^a2*...*pk^ak
phi(n)=p1^a1*(1-1/p1)*p2^a2*(1-1/p2)*...*pk^ak*(1-1/pk)=n*(1-1/p1)*(1-1/p2)*...*(1-1/pk)
1 int euler(int n) 2 { 3 int ans = n; 4 for (int i = 2; i * i <= n; i++) 5 if (n % i == 0) 6 { 7 ans = ans / i * (i - 1); 8 while (n % i == 0) 9 n /= i; 10 } 11 if (n > 1) 12 ans = ans / n * (n - 1); 13 return ans; 14 }
1 void phi_table(int *phi, int *f, int *p, int &pn) 2 { 3 phi[1] = 1; 4 pn = 0; 5 int k; 6 for (int i = 2; i < N; i++) 7 { 8 if (!f[i]) 9 { 10 p[pn++] = f[i] = i; 11 phi[i] = i - 1; 12 } 13 for (int j = 0; j < pn && (k = p[j] * i) < N; j++) 14 { 15 f[k] = p[j]; 16 if (f[i] == p[j]) 17 { 18 phi[k] = phi[i] * p[j]; 19 break; 20 } 21 else 22 phi[k] = phi[i] * (p[j] - 1); 23 } 24 } 25 //for(int i=2;i<=N;i++) phi[i]+=phi[i-1];//累加 26 }
1 void prime(int &pn, int *p, bool *vis) 2 { 3 pn = 0; 4 for (int i = 2; i < N; i++) 5 { 6 if (!vis[i]) 7 p[pn++] = i; 8 for (int j = 0; j < pn && i * p[j] < N; j++) 9 { 10 vis[i * p[j]] = 1; 11 if (i % p[j] == 0) 12 break; 13 } 14 } 15 }
1 int vis[N / 31], pn, p[N / 10]; 2 inline bool judge(int x) { return vis[x >> 5] & (1 << (x & 31)); } 3 inline void setbit(int x) { vis[x >> 5] |= 1 << (x & 31); } 4 void init() 5 { 6 pn = 0; 7 memset(vis, 0, sizeof(vis)); 8 for (int i = 2; i < M; i++) 9 { 10 if (!judge(i)) 11 p[pn++] = i; 12 for (int j = 0; p[j] * i < M; j++) 13 { 14 setbit(i * p[j]); 15 if (i % p[j] == 0) 16 break; 17 } 18 } 19 }
1 int prime(int a, int b) 2 { 3 memset(vis, 1, sizeof(vis)); 4 for (int i = 0; i < pn && (LL)p[i] * p[i] <= b; i++) 5 { 6 int x = a / p[i] + (a % p[i] != 0); 7 if (x == 1) 8 x++; 9 for (LL j = (LL)x * p[i]; j <= b; j += p[i]) 10 vis[j - a] = 0; //,printf("%d ",j); 11 } 12 int ans = 0; 13 for (LL i = 0; i + a <= b; i++) 14 ans += vis[i]; 15 if (a == 1) 16 ans--; 17 return ans; 18 }
1 LL cal(LL p, LL n) 2 { 3 p %= mod; 4 if (p == 1) 5 return (n + 1) % mod; 6 LL inv = quick_mod(p - 1, mod - 2); 7 LL t = quick_mod(p, n); 8 if (t * p % mod != 1) 9 return (t * p % mod - 1) * inv % mod; 10 return (t - 1) * inv % mod * t % mod; 11 }
1 int log_mod(int a, int b) 2 { 3 int m, v, e = 1; 4 m = (int)sqrt(mod + 0.5); 5 v = Inv(quick_mod(a, m)); 6 map<int, int> x; 7 x[1] = 0; 8 for (int i = 1; i < m; i++) 9 { 10 e = (LL)e * a % mod; 11 if (!x.count(e)) 12 x[e] = i; 13 } 14 15 for (int i = 0; i < m; i++) 16 { 17 if (x.count(b)) 18 return i * m + x[b]; 19 b = (LL)b * v % mod; 20 } 21 return -1; 22 }
1 void gcd(LL a, LL b, LL &d, LL &x, LL &y) 2 { 3 if (!b) 4 { 5 d = a; 6 x = 1; 7 y = 0; 8 } 9 else 10 { 11 gcd(b, a % b, d, y, x); 12 y -= a / b * x; 13 } 14 }
1 LL equation(LL a, LL b, LL r, LL &x, LL &y) 2 { 3 LL g; 4 gcd(a, b, g, x, y); //求出 ax+by==g 5 if (r % g) 6 return 0; //无解 7 x *= r / g; 8 y *= r / g; 9 return g > 0 ? g : -g; //返回 gcd 10 }
1 int a, b, c, x0, x2, y0, y2; 2 LL solve(LL a, LL b, LL c, LL x0, LL x2, LL y0, LL y2) 3 { 4 if (a < 0 && b < 0) 5 a = -a, b = -b, c = -c; 6 if (!b || a < 0 && b > 0) 7 swap(a, b), swap(x0, y0), swap(x2, y2); 8 if (!a && !b) 9 { 10 if (c) 11 return 0; 12 return (x2 - x0 + 1LL) * (y2 - y0 + 1LL); 13 } 14 if (!a) 15 { 16 if (c % b || y0 > -c / b || y2 < -c / b) 17 return 0; 18 return x2 - x0 + 1; 19 } 20 LL g, x, y; 21 g = equation(a, b, -c, x, y); 22 if (!g) 23 return 0; 24 25 LL aa = abs(a / g), bb = abs(b / g); 26 x0 += ((x % bb + bb) % bb - (x0 % bb + bb) % bb + bb) % bb; 27 x2 -= ((x2 % bb + bb) % bb - (x % bb + bb) % bb + bb) % bb; 28 y0 += ((y % aa + aa) % aa - (y0 % aa + aa) % aa + aa) % aa; 29 y2 -= ((y2 % aa + aa) % aa - (y % aa + aa) % aa + aa) % aa; 30 if (x0 > x2 || y0 > y2) 31 return 0; 32 x0 = (-c - a * x0) / b; 33 //a*x0注意溢出 34 x2 = (-c - a * x2) / b; 35 if (b > 0) 36 swap(x0, x2); 37 x0 = max(x0, y0); 38 x2 = min(x2, y2); 39 if (x0 > x2) 40 return 0; 41 return (x2 - x0) / aa + 1; 42 }
1 LL multi_equation(int n, int *r, int *m) 2 { 3 LL x, y, g, lcm, ans; 4 lcm = m[0]; 5 ans = r[0]; 6 for (int i = 1; i < n; i++) 7 { 8 g = equation(lcm, -m[i], r[i] - ans, x, y); 9 if (!g) 10 return -1; //无解 11 ans += x * lcm; 12 lcm *= m[i] / g; 13 ans %= lcm; 14 //注意溢出 15 } 16 if (ans % lcm == 0) 17 return lcm; //结果为 0时答案为 lcm 18 return (ans + lcm) % lcm; 19 }
1 LL Inv(LL a, LL n) //O(logn) 2 { 3 LL g, x, y; 4 gcd(a, n, g, x, y); 5 return g == 1 ? (x + n) % n : -1; 6 }
//逆元求(a^b-c)/d%p时当 a^b-c==0时(a^b-c)*inv(d)==0会出错,需特殊处理
//逆元,a/b%mod=a*b^-1%mod，f(x)=inv(x,mod)或 f(x)=quick_mod(x,mod-2,mod)，mod需是质数
//若 a,m互质,r=a/b%m=a*inv(b)%m,否则 r=a/b%m=a%(m*b)/b 可以算 a%(m*b)但算不了 a/b时用
//mod为素数线性求逆元:inv(x)=-(mod/x)*inv(mod%x)=(mod– mod/x)*inv(mod%x)%mod
1 LL quick_mod(LL t, LL n) 2 { 3 t %= mod; 4 LL ans = 1; 5 while (n) 6 { 7 if (n & 1) 8 ans = ans * t % mod; 9 t = t * t % mod; 10 n >>= 1; 11 } 12 return ans; 13 }
1 LL fac[N]; 2 LL Inv(LL x) 3 { 4 return quick_mod(x, mod - 2); 5 } 6 void init() 7 { 8 //要先执行 9 fac[0] = 1; 10 for (int i = 1; i < N; i++) 11 fac[i] = fac[i - 1] * i % mod; 12 //阶乘模 13 } 14 //若 mod不为质数且 n*m较小可以考虑打 n*m的 C[n][m]二维表 15 LL Cnm(int n, int m) 16 { 17 if (n < m || n < 0 || m < 0) 18 return 0; 19 return fac[n] * Inv(fac[m] * fac[n - m] % mod) % mod; 20 }
1 LL Lucas(LL n, LL m) 2 { 3 if (n < m || n < 0 || m < 0) 4 return 0; 5 LL ans = 1; 6 while (n && m) 7 { 8 ans = ans * Cnm(n % mod, m % mod) % mod; //这儿 Cnm的求逆不可先处理，会超时 9 n /= mod; 10 m /= mod; 11 } 12 return ans; 13 } 14 LL Lucas(LL n, LL m) 15 //递归 16 { 17 if (m == 0) 18 return 1; 19 else 20 return (Cnm(n % mod, m % mod) * Lucas(n / mod, m / mod)) % mod; 21 }
/*首先给出这个 Lucas定理：
A、B是非负整数，p是质数。AB写成 p进制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。
则组合数 C(A,B)与 C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0]) mod p同余
即：Lucas(n,m,p)=c(n%p,m%p)*Lucas(n/p,m/p,p) */
//求 C(n,m)%mod mod最大为 10^5。a,b可以很大！
1 //对每个数分解质因子和次幂 2 int pn, p[N], vis[N], d[N][10], e[N][10], dn[N]; 3 void init() 4 { 5 pn = 1; 6 memset(vis, 0, sizeof(vis)); 7 memset(e, 0, sizeof(e)); 8 memset(dn, 0, sizeof(dn)); 9 for (int i = 2; i < N; i++) 10 if (!vis[i]) 11 { 12 p[pn] = i; 13 for (int j = i; j < N; j += i) 14 { 15 vis[j] = 1; 16 int &n = dn[j]; 17 d[j][n] = i; 18 int x = j; 19 while (x % i == 0) 20 x /= i, e[j][n]++; 21 n++; 22 } 23 vis[i] = pn++; 24 } 25 } 26 27 int c[N]; 28 inline void add(int n, int x) 29 { 30 if (!x) 31 return; 32 for (int i = 0; i < dn[n]; i++) 33 { 34 int id = vis[d[n][i]]; 35 c[id] += x * e[n][i]; 36 } 37 } 38 //x个组合数相乘求模,mod可不为质数,n不能太大,复杂度 x*log(n)+n*log(p) 39 int bit[N]; 40 inline int sum(int x) 41 { 42 int res = 0; 43 while (x > 0) 44 res += bit[x], x -= x & -x; 45 return res; 46 } 47 inline void update(int x, int y, int d) 48 { 49 if (!d) 50 return; 51 while (x <= n) 52 bit[x] += d, x += x & -x; 53 y++; 54 while (y <= n) 55 bit[y] -= d, y += y & -y; 56 } 57 int vv[n]; 58 inline void Cnm(int n, int m) 59 { 60 vv[n]++; 61 vv[m]--; 62 vv[n - m]--; 63 } 64 LL cal() 65 { 66 LL ans = 1; 67 for (int i = 2; i <= n; i++) 68 update(1, i, vv[i]); 69 for (int i = 2; i <= n; i++) 70 add(i, sum(i)); 71 for (int i = 1; i < pn && p[i] <= n; i++) 72 if (c[i]) 73 ans = ans * quick_mod(p[i], c[i]) % mod; 74 return ans; 75 } 76 77 //单个组合数取 mod,mod可不为质数 复杂度 n*log(p) 78 LL Cnm(int n, int m) 79 { 80 if (n < m || n < 0 || m < 0) 81 return 0; 82 m = min(m, n - m); 83 LL ans = 1; 84 memset(c, 0, sizeof(c)); 85 for (int i = n, j = m; j > 0; i--, j--) 86 { 87 add(i, 1); 88 add(j, -1); 89 } 90 for (int i = 1; i < pn; i++) 91 if (c[i]) 92 ans = ans * quick_mod(p[i], c[i], mod) % mod; 93 return ans; 94 } 95 int c[N], vv[N]; 96 inline void add(int n, int x) 97 { 98 if (!x) 99 return; 100 for (int i = 0; i < dn[n]; i++) 101 { 102 int id = vis[d[n][i]]; 103 c[id] += x * e[n][i]; 104 } 105 }
1 int c[N][N], fac[N], part[N][N]; 2 void init() 3 { 4 memset(c, 0, sizeof(c)); 5 c[0][0] = fac[0] = 1; 6 for (int i = 1; i < N; i++) //杨辉三角 Cnm递推 7 { 8 fac[i] = (LL)fac[i - 1] * i % mod; 9 c[i][0] = 1; 10 for (int j = 1; j <= i; j++) 11 c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod; 12 } 13 //集合划分,part[i][j]表示 i个不同的元素划分成 j个相同组(i个不同球放到 j个相同盒子,不能有空盒, 14 //可以有空盒的为前缀和),递推式 S(n,m)=S(n-1,m-1)+m*S(n-1,m) 15 memset(part, 0, sizeof(part)); 16 part[1][1] = 1; 17 for (int i = 2; i < N; i++) 18 for (int j = 1; j <= i; j++) 19 part[i][j] = (part[i - 1][j] * j + part[i - 1][j - 1]) % mod; 20 }
1 typedef unsigned int UI; 2 const int HASH=100007; 3 char e[N][20],w[N][20]; 4 struct Hash 5 { 6 int *head, *next, n; 7 Hash() 8 { 9 head = new int[HASH]; 10 next = new int[N]; 11 n = 1; 12 memset(head, 0, sizeof(head)); 13 } 14 ~Hash() 15 { 16 delete[] head; 17 delete[] next; 18 } 19 UI hash(char *s) 20 { 21 UI ans = 0; 22 for (int i = 0; s[i]; i++) 23 ans = s[i] + ans * 101; 24 return ans % HASH; 25 } 26 int hash(int s) { return s % HASH; } 27 bool cmp(char *a, char *b) { return !strcmp(a, b); } 28 bool cmp(int a, int b) { return a == b; } 29 bool insert(int s) 30 { 31 int h = hash(e[s]); 32 int u = head[h]; 33 while (u) 34 { 35 //当插入元素中不会有重复时可不要此循环 36 if (cmp(e[u], e[s])) 37 return 0; 38 u = next[u]; 39 } 40 next[s] = head[h]; 41 head[h] = s; 42 return 1; 43 } 44 int find(char *s) 45 { 46 int h = hash(s); 47 int u = head[h]; 48 while (u) 49 { 50 if (cmp(e[u], s)) 51 return u; 52 u = next[u]; 53 } 54 return 0; 55 } 56 };
1 const LL base=1e18; 2 const LL MOD=1e9; 3 struct int128 4 { 5 LL a, b; 6 //a为低位 7 int128(LL x = 0, LL y = 0) 8 { 9 a = x; 10 b = y; 11 } 12 int128 operator+(const int128 &x) const 13 { 14 return int128(a + x.a, b + x.b); 15 } 16 int128 operator-(const int128 &x) const 17 { 18 return int128(a - x.a, b - x.b); 19 } 20 int128 operator*(const int128 &x) const //相乘结果范围为[-(2^63)*1e18,(2^63-1)*1e18] 21 { 22 LL x0 = a % MOD, x1 = a / MOD; 23 LL y0 = x.a % MOD, y1 = x.a / MOD; 24 int128 c; 25 c.a = x0 * y0; 26 c.b = x1 * y1; 27 c.a += x0 * y1 % MOD * MOD; 28 c.b += x0 * y1 / MOD; 29 c.a += x1 * y0 % MOD * MOD; 30 c.b += x1 * y0 / MOD; 31 return c; 32 } 33 int128 operator/(const LL &x) const 34 { 35 LL ans, r, t = a, e = base / 10; 36 ans = b / x; 37 r = b % x; 38 for (int i = 17; i >= 0; i--, e /= 10) 39 { 40 ans = ans * 10 + (r * 10 + t / e) / x; 41 r = (r * 10 + t / e) % x; 42 t %= e; 43 } 44 return int128(ans); 45 } 46 int128 operator%(const LL &x) const 47 { 48 LL ans, t = a, e = base / 10; 49 ans = b % x; 50 for (int i = 17; i >= 0; i--, e /= 10) 51 { 52 ans = (ans * 10 + t / e) % x; 53 t %= e; 54 } 55 return int128(ans); 56 } 57 }; 58 59 60 void print(int128 x) 61 { 62 if (x.b) 63 printf("%lld%018lld", x.b, x.a); 64 else 65 printf("%lld", x.a); 66 }
1 const int base=10; 2 int largediv(char*a,char*b,char*ans) 3 { 4 int j = 0, la = strlen(a), lb = strlen(b), l = 0; 5 while (la - j >= lb) 6 { 7 int t = 0; 8 while ((j > 0 && a[j - 1] != '0') || strncmp(a + j, b, lb) >= 0) 9 { 10 t++; 11 for (int i = lb - 1; i >= 0; i--) 12 { 13 a[i + j] -= b[i] - '0'; 14 if (a[i + j] < '0') 15 { 16 a[i + j] += base; 17 a[i + j - 1]--; 18 } 19 } 20 } 21 j++; 22 ans[l++] = t + '0'; 23 if (l == 0 && ans[l] == '0') 24 l--; 25 } 26 if (!l) 27 ans[l++] = '0'; 28 ans[l] = 0; 29 int x = 0; 30 while (x < la && a[x] == '0') 31 x++; 32 if (x == la) 33 x--; 34 return x; 35 }
1 struct Fraction 2 { 3 LL num; 4 LL den; 5 Fraction(LL num = 0, LL den = 1) 6 { 7 if (den < 0) 8 { 9 num = -num; 10 den = -den; 11 } 12 assert(den != 0); 13 LL g = __gcd(abs(num), den); 14 this->num = num / g; 15 this->den = den / g; 16 } 17 Fraction operator+(const Fraction &x) const 18 { 19 return Fraction(num * x.den + den * x.num, den * x.den); 20 } 21 Fraction operator-(const Fraction &x) const 22 { 23 return Fraction(num * x.den - den * x.num, den * x.den); 24 } 25 Fraction operator*(const Fraction &x) const 26 { 27 return Fraction(num * x.num, den * x.den); 28 } 29 Fraction operator/(const Fraction &x) const 30 { 31 return Fraction(num * x.den, den * x.num); 32 } 33 bool operator<(const Fraction &x) const 34 { 35 return num * x.den < den * x.num; 36 } 37 bool operator==(const Fraction &x) const 38 { 39 return num * x.den == den * x.num; 40 } 41 };
1 struct bigint 2 { 3 int *a; 4 //a[0]存长度 5 static const int maxlen = 1e4 + 10; 6 static const int basel = 8; 7 static const int base = 1e9; 8 bigint() { a = new int[maxlen]; } 9 ~bigint() { delete[] a; } 10 //输出大整数 a,base不一样时记得改 04d 11 void print() 12 { 13 int n = a[0]; 14 printf("%d", a[--n]); 15 for (--n; n; n--) 16 printf("%09d", a[n]); 17 printf(" "); 18 } 19 //将正序的字符串 x转为 10^(basel+1)进制逆序大整数 a 20 bigint &trans(char *x) 21 { 22 int lx = strlen(x); 23 int &la = a[0] = 1; 24 for (int i = lx - 1; i >= 0;) 25 { 26 int t = 0, k = i - (basel < i ? basel : i); 27 for (int j = k; j <= i; j++) 28 t = t * 10 + x[j] - '0'; 29 i = k - 1; 30 a[la++] = t; 31 } 32 return *this; 33 } 34 bool equal(int x) { return a[0] == 2 && a[1] == x; } 35 //<base的整数转为大整数 36 bigint &intTobig(int x) 37 { 38 a[0] = 2; 39 a[1] = x; 40 return *this; 41 } 42 bigint ©(bigint & b) 43 { 44 memcpy(a, b.a, sizeof(a[0]) * b.a[0]); 45 return *this; 46 } 47 void swap(bigint & b) 48 { 49 int *tmp = a; 50 a = b.a; 51 b.a = tmp; 52 } 53 bigint &add(bigint & tb, bigint & tans) 54 { 55 int *b = tb.a, *ans = tans.a; 56 int l = max(a[0], b[0]); 57 memset(ans, 0, sizeof(a[0]) * (l + 1)); 58 for (int i = 1; i < l; i++) 59 { 60 if (i >= a[0]) 61 a[i] = 0; 62 if (i >= b[0]) 63 b[i] = 0; 64 ans[i] += a[i] + b[i]; 65 ans[i + 1] += ans[i] / base; 66 ans[i] %= base; 67 } 68 ans[0] = (ans[l] ? l + 1 : l); 69 return tans; 70 } 71 bigint &minus(bigint & tb, bigint & tans) 72 { 73 int *b = tb.a, *ans = tans.a; 74 memset(ans, 0, sizeof(int) * (a[0] + 1)); 75 for (int i = 1; i < a[0]; i++) 76 { 77 if (i >= b[0]) 78 b[i] = 0; 79 ans[i] += a[i] - b[i]; 80 if (ans[i] < 0) 81 { 82 ans[i] += base; 83 ans[i + 1]--; 84 } 85 } 86 ans[0] = a[0]; 87 while (ans[0] > 2 && !ans[ans[0] - 1]) 88 ans[0]--; 89 return tans; 90 } 91 bigint &mul(bigint & tb, bigint & tans) 92 { 93 int *b = tb.a, *ans = tans.a; 94 memset(ans, 0, sizeof(int) * (a[0] + b[0])); 95 for (int i = 1; i < a[0]; i++) 96 { 97 for (int j = 1; j < b[0]; j++) 98 { 99 LL t = ans[i + j - 1] + (LL)a[i] * b[j]; 100 ans[i + j] += t / base; 101 ans[i + j - 1] = t % base; 102 } 103 } 104 int &l = ans[0]; 105 l = a[0] + b[0] - 2; 106 107 while (l > 1 && !ans[l]) 108 l--; 109 l++; 110 return tans; 111 } 112 //大整数除法或求模,被除数<=Base 113 int div(int b, bigint &tans) 114 { 115 int *ans = tans.a; 116 ans[0] = 1; 117 ans[ans[0]++] = 0; 118 LL sum = 0, first = 1; 119 for (int i = a[0] - 1; i > 0; i--) 120 { 121 sum = sum * base + a[i]; 122 if (sum < b && first) 123 continue; 124 if (first) 125 { 126 first = 0; 127 ans[0] = i + 1; 128 } 129 ans[i] = sum / b; 130 sum %= b; 131 } 132 return sum; 133 } 134 bool less(bigint & x) //this<x 135 { 136 if (x.a[0] != a[0]) 137 return a[0] < x.a[0]; 138 for (int i = a[0] - 1; i > 0; i--) 139 if (a[i] != x.a[i]) 140 return a[i] < x.a[i]; 141 return 0; 142 } 143 };
1 int d[2000],dn,c[2000]; 2 void fac(int n)//找出所有约数 3 { 4 dn = 0; 5 for (int i = 1; i * i <= n; i++) 6 if (n % i == 0) 7 { 8 d[dn++] = i; 9 if (i * i != n) 10 d[dn++] = n / i; 11 } 12 sort(d, d + dn); 13 } 14 LL polya(int n) 15 { 16 fac(n); 17 memset(c, 0, sizeof(c)); 18 for (int i = 0; i < dn; i++) //容斥求所有约数的欧拉函数 19 { 20 c[i] += d[i]; 21 for (int j = i + 1; j < dn; j++) 22 if (d[j] % d[i] == 0) 23 c[j] -= c[i]; 24 } 25 LL ans = 0; 26 for (int i = 0; i < dn; i++) 27 { 28 int x = n / d[i]; 29 ans = (ans + (LL)c[i] * mar_quick(x, 2, 1)) % mod; //计算 x 个循环节的置换有多少种方案,polya 定理中 30 为 m ^ x 31 } 32 ans = (ans * Inv(n)) % mod; 33 return ans; 34 }
1 int d[2000],phi[2000],dn,p[20],pn,e[20]; 2 void facdiv(int n) //质因分解成 n=p[0]^e[0] * p[1]^e[1] * ... *p[pn-1]^e[pn-1] 3 { 4 pn = 0; 5 //for(int j=0,i=prime[0];i*i<=n;i=prime[++j]) if(n%i==0) 6 //筛出质数表加速质因分解 7 for (int i = 2; i * i <= n; i++) 8 if (n % i == 0) 9 { 10 p[pn] = i; 11 e[pn] = 0; 12 while (n % i == 0) 13 n /= i, e[pn]++; 14 pn++; 15 } 16 if (n != 1) 17 p[pn] = n, e[pn++] = 1; 18 }
1 void GetDivisorPhi() //根据上面的基因分解算出所有约数的欧拉函数(与 d[i]互质的数的个数) 2 { 3 dn = 1; 4 d[0] = 1; 5 phi[0] = 1; 6 for (int i = 0; i < pn; i++) 7 { 8 int num = dn; 9 for (int j = 0; j < num; j++) 10 { 11 phi[dn] = phi[j] * (p[i] - 1); 12 d[dn] = d[j] * p[i]; 13 dn++; 14 for (int k = 2; k <= e[i]; k++) 15 { 16 phi[dn] = phi[dn - 1] * p[i]; 17 d[dn] = d[dn - 1] * p[i]; 18 dn++; 19 } 20 } 21 } 22 }
1 const int S=20;//随机算法判定次数，S越大，判错概率越小 2 //计算 (a*b)%c. a,b都是 LL的数，直接相乘可能溢出的 3 // a,b,c <2^63 4 LL multi_mod( LL x , LL y ,LL mod) 5 { 6 return (x * y - (LL)(x / (long double)mod * y + 1e-3) * mod + mod) % mod; 7 } 8 //快速幂，计算 x^n %c 9 LL quick_mod(LL t,LL n,LL mod) 10 { 11 t %= mod; 12 LL ans = 1; 13 while (n) 14 { 15 if (n & 1) 16 ans = multi_mod(ans, t, mod); 17 t = multi_mod(t, t, mod); 18 n >>= 1; 19 } 20 return ans; 21 } 22 23 24 //以 a为基,n-1=x*2^t 25 a^(n-1)=1(mod n) 验证 n是不是合数 26 //一定是合数返回 true,不一定返回 false 27 bool check(LL a,LL n,LL x,LL t) 28 { 29 LL ans = quick_mod(a, x, n); 30 LL last = ans; 31 for (int i = 1; i <= t; i++) 32 { 33 ans = multi_mod(ans, ans, n); 34 if (ans == 1 && last != 1 && last != n - 1) 35 return true; //合数 36 last = ans; 37 } 38 if (ans != 1) 39 return true; 40 return false; 41 } 42 // Miller_Rabin()算法素数判定 43 //是素数返回 true.(可能是伪素数，但概率极小) 44 //合数返回 false; 45 bool Miller_Rabin(LL n) 46 { 47 if (n < 2) 48 return false; 49 if (n == 2) 50 return true; 51 if ((n & 1) == 0) 52 return false; //偶数 53 LL x = n - 1; 54 LL t = 0; 55 while ((x & 1) == 0) 56 { 57 x >>= 1; 58 t++; 59 } 60 for (int i = 0; i < S; i++) 61 { 62 LL a = rand() % (n - 1) + 1; 63 if (check(a, n, x, t)) 64 return false; //合数 65 } 66 return true; 67 }
1 LL fac[100];//质因数分解结果（刚返回时是无序的） 2 int tol;//质因数的个数。数组小标从 0开始 3 LL gcd(LL a,LL b) 4 { 5 if (a == 0) 6 return 1; 7 if (a < 0) 8 return gcd(-a, b); 9 while (b) 10 { 11 LL t = a % b; 12 a = b; 13 b = t; 14 } 15 return a; 16 } 17 18 19 LL Pollard_rho(LL x,LL c) 20 { 21 LL i = 1, k = 2; 22 LL x0 = rand() % x; 23 LL y = x0; 24 while (1) 25 { 26 i++; 27 x0 = (multi_mod(x0, x0, x) + c) % x; 28 LL d = gcd(y - x0, x); 29 if (d != 1 && d != x) 30 return d; 31 if (y == x0) 32 return x; 33 if (i == k) 34 y = x0, k += k; 35 } 36 } 37 //对 n进行素因子分解，分解后的因子不是有序的，要 sort 38 void findfac(LL n) 39 { 40 if (n == 1) 41 return; 42 if (Miller_Rabin(n)) //素数 43 { 44 fac[tol++] = n; 45 return; 46 } 47 LL p = n; 48 while (p >= n) 49 p = Pollard_rho(p, rand() % (n - 1) + 1); 50 findfac(p); 51 findfac(n / p); 52 }
1 LL multi_mod(LL a,LL b,LL mod)//a,b传过来的要是正数 2 { 3 LL ans = 0; 4 a %= mod; 5 b %= mod; 6 while (b) 7 { 8 if (b & 1) 9 ans = ans + a; 10 if (ans >= mod) 11 ans -= mod; 12 a <<= 1; 13 if (a >= mod) 14 a -= mod; 15 b >>= 1; 16 } 17 return ans; 18 }
1 LL multi_mod( LL x , LL y ,LL mod) 2 { 3 return (x * y - (LL)(x / (long double)mod * y + 1e-3) * mod + mod) % mod; 4 }
1 //multi_mod 中过多的传参会耗费大量时间,不爆 long long 的话最好写成 a*b%mod 2 //如果精度超过 double 可以用 NTT 求多个 P,再用中国剩余定理算回去 3 //f(i)=sigma{a[j]*b[i-j],0<=i<n} 4 //const LL P = ( 1LL << 55 ) * 5 + 1 ; 5 //const LL G = 6 ; 6 const int P =479<<21|1; 7 //const int P1=119<<23|1;//费马质数,P-1 能整除 2^n 8 const int G = 3; //原根,最小的数满足 G^(P-1)==1 9 const int NUM = 30; 10 LL wn[NUM]; 11 LL x[N*4],y[N*4]; 12 void GetWn() 13 { 14 for (int i = 0; i < NUM; i++) 15 { 16 int t = 1 << i; 17 wn[i] = quick_mod(G, (P - 1) / t); 18 } 19 } 20 void NTT(LL *a,int len,bool rev) 21 { 22 for (int i = 1, j, t, k; i < len; i++) 23 { 24 for (k = len >> 1, t = i, j = 0; k; k >>= 1, t >>= 1) 25 j = j << 1 | t & 1; 26 if (i < j) 27 swap(a[i], a[j]); 28 } 29 for (int s = 2, ds = 1, id = 1; s <= len; ds = s, s <<= 1, id++) 30 { 31 for (int k = 0; k < len; k += s) 32 { 33 LL w = 1, t; 34 for (int i = k; i < k + ds; i++) 35 { 36 a[i + ds] = (a[i] - (t = multi_mod(a[i + ds], w)) + P) % P; 37 a[i] = (a[i] + t) % P; 38 w = multi_mod(w, wn[id]); 39 } 40 } 41 } 42 if (rev) 43 { 44 for (int i = 1; i<len>> 1; i++) 45 swap(a[i], a[len - i]); 46 LL inv = quick_mod(len, P - 2); 47 for (int i = 0; i < len; i++) 48 a[i] = multi_mod(a[i], inv); 49 } 50 } 51 void Conv(LL *a, LL *b, int n) 52 { 53 NTT(a, n, 0); 54 NTT(b, n, 0); 55 for (int i = 0; i < n; i++) 56 a[i] = multi_mod(a[i], b[i]); 57 NTT(a, n, 1); 58 }
1 const double pi=acos(-1); 2 struct cplx 3 { 4 //复数类/多项式/大整数 5 double r, i; 6 cplx(double r = 0, double i = 0) : r(r), i(i) {} 7 cplx operator+(const cplx &a) const 8 { 9 return cplx(r + a.r, i + a.i); 10 } 11 cplx operator-(const cplx &a) const 12 { 13 return cplx(r - a.r, i - a.i); 14 } 15 cplx operator*(const cplx &a) const 16 { 17 return cplx(r * a.r - i * a.i, r * a.i + i * a.r); 18 } 19 }; 20 //快速傅里叶变换 FFT(nlogn) 21 //len = 2^M,reverse F[i] with F[j] j为 i二进制反转 22 void rader(cplx *f,int len) 23 { 24 int j = len >> 1; 25 for (int i = 1; i < len - 1; ++i) 26 { 27 if (i < j) 28 swap(f[i], f[j]); // reverse 29 int k = len >> 1; 30 while (j >= k) 31 { 32 j -= k; 33 k >>= 1; 34 } 35 if (j < k) 36 j += k; 37 } 38 } 39 //做 FFT,len必须为 2^k形式，on==1时是 DFT，on==-1时是 IDFT 40 void FFT(cplx *f,int len,int t) 41 { 42 rader(f, len); 43 for (int h = 2; h <= len; h <<= 1) 44 { 45 cplx wn(cos(-t * 2 * pi / h), sin(-t * 2 * pi / h)); 46 for (int j = 0; j < len; j += h) 47 { 48 cplx e(1, 0); //旋转因子 49 for (int k = j; k < j + h / 2; ++k) 50 { 51 cplx u = f[k]; 52 53 cplx v = e * f[k + h / 2]; 54 f[k] = u + v; 55 f[k + h / 2] = u - v; 56 e = e * wn; 57 } 58 } 59 } 60 if (t == -1) 61 for (int i = 0; i < len; ++i) 62 f[i].r /= len; //IDFT 63 } 64 //传入 a，b之前需将 a，b的长度处理为 len=2^k,a，b数组需开到原长度的 3-4倍,变换结果为 a数组 65 void Conv(cplx *a,cplx *b,int len) //求卷积 66 { 67 FFT(a, len, 1); 68 FFT(b, len, 1); 69 for (int i = 0; i < len; ++i) 70 a[i] = a[i] * b[i]; 71 FFT(a, len, -1); 72 }
1 const int N = 5e6+2; 2 bool np[N]; 3 int p[N],pi[N]; 4 int getprime() 5 { 6 //pi[i]表示不超过 i的质数个数，p存素数 7 int cnt = 0; 8 np[0] = np[1] = true; 9 pi[0] = pi[1] = 0; 10 for (int i = 2; i < N; ++i) 11 { 12 if (!np[i]) 13 p[++cnt] = i; 14 for (int j = 1; j <= cnt && i * p[j] < N; ++j) 15 { 16 np[i * p[j]] = true; 17 } 18 pi[i] = cnt; 19 } 20 return cnt; 21 } 22 const int M = 7; 23 const int PM = 2*3*5*7*11*13*17; 24 int phi[PM+1][M+1],sz[M+1]; 25 void init() 26 { 27 getprime(); 28 sz[0] = 1; 29 for (int i = 0; i <= PM; ++i) 30 phi[i][0] = i; 31 for (int i = 1; i <= M; ++i) 32 { 33 sz[i] = p[i] * sz[i - 1]; 34 for (int j = 1; j <= PM; ++j) 35 { 36 phi[j][i] = phi[j][i - 1] - phi[j / p[i]][i - 1]; 37 } 38 } 39 } 40 41 42 int sqrt2(LL x) 43 { 44 LL r = (LL)sqrt(x - 0.1); 45 while (r * r <= x) 46 ++r; 47 return int(r - 1); 48 } 49 int sqrt3(LL x) 50 { 51 LL r = (LL)cbrt(x - 0.1); 52 while (r * r * r <= x) 53 ++r; 54 return int(r - 1); 55 } 56 LL getphi(LL x,int s) 57 { 58 if (s == 0) 59 return x; 60 if (s <= M) 61 return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s]; 62 if (x <= p[s] * p[s]) 63 return pi[x] - s + 1; 64 if (x <= p[s] * p[s] * p[s] && x < N) 65 { 66 int s2x = pi[sqrt2(x)]; 67 LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2; 68 for (int i = s + 1; i <= s2x; ++i) 69 { 70 ans += pi[x / p[i]]; 71 } 72 return ans; 73 } 74 return getphi(x, s - 1) - getphi(x / p[s], s - 1); 75 } 76 LL getpi(LL x) 77 { 78 //返回小于 x 的素数个数 79 if (x < N) 80 return pi[x]; 81 LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1; 82 for (int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) 83 ans -= getpi(x / p[i]) - i + 1; 84 return ans; 85 } 86 LL lehmer_pi(LL x) 87 { 88 if (x < N) 89 return pi[x]; 90 int a = (int)lehmer_pi(sqrt2(sqrt2(x))); 91 int b = (int)lehmer_pi(sqrt2(x)); 92 int c = (int)lehmer_pi(sqrt3(x)); 93 LL sum = getphi(x, a) + LL(b + a - 2) * (b - a + 1) / 2; 94 for (int i = a + 1; i <= b; i++) 95 { 96 LL w = x / p[i]; 97 sum -= lehmer_pi(w); 98 if (i > c) 99 continue; 100 LL lim = lehmer_pi(sqrt2(w)); 101 for (int j = i; j <= lim; j++) 102 { 103 sum -= lehmer_pi(w / p[j]) - (j - 1); 104 } 105 } 106 return sum; 107 108 109 } 110 LL getans(LL x) 111 { // x < 1e11 112 LL ans = 0; 113 //返回相乘不大于 x的质数对数 114 for (int i = 1, ed = pi[sqrt2(x - 1)]; i <= ed; ++i) 115 ans += lehmer_pi(x / p[i]) - i; 116 return ans; 117 }
1 void swap(int &a,int &b) 2 { 3 a ^= b; 4 b ^= a; 5 a ^= b; 6 }
1 struct node 2 { 3 int i, w; 4 bool operator<(const node &a) const 5 { 6 return w > a.w; 7 } 8 };
1 int prim() 2 { 3 //O(nlogn) 4 memset(d, 0x3f, sizeof(d)); 5 memset(vis, 0, sizeof(vis)); 6 d[1] = 0; 7 priority_queue<node> q; 8 q.push(node{1, 0}); 9 while (!q.empty()) 10 { 11 node u = q.top(); 12 q.pop(); 13 if (vis[u.i]) 14 continue; 15 vis[u.i] = 1; 16 for (vector<int>::iterator i = v[u.i].begin(), j = w[u.i].begin(); i != v[u.i].end(); i++, j++) 17 { 18 if (vis[*i] || d[*i] <= *j) 19 continue; 20 d[*i] = *j; 21 q.push(node{*i, d[*i]}); 22 } 23 } 24 int sum = 0; 25 for (int i = 1; i <= n; i++) 26 sum += d[i]; 27 return sum; 28 } 29 30 31 int prim() 32 { 33 //O(n^2) 34 memset(d, 0x3f, sizeof(d)); 35 memset(vis, 0, sizeof(vis)); 36 d[1] = 0; 37 for (int i = 2; i <= n; i++) 38 { 39 int x = -1; 40 for (int j = 1; j <= n; j++) 41 if (!vis[j] && (x == -1 || d[j] < d[x])) 42 x = j; //j要未作为过起点 43 vis[x] = 1; 44 for (vector<int>::iterator j = v[x].begin(), k = w[x].begin(); j != v[x].end(); j++, k++) 45 if (!vis[*j] && *k < d[*j]) 46 d[*j] = *k; 47 } 48 int sum = 0; 49 for (int i = 1; i <= n; i++) 50 sum += d[i]; 51 return sum; 52 }
1 struct maxMacth 2 { 3 int n; 4 int vm[N], um[N]; 5 bool vis[N]; 6 vector<int> g[N]; 7 int dx[N], dy[N], dis; 8 void init(int num) 9 { 10 n = num; 11 memset(vm, -1, sizeof(vm)); 12 memset(um, -1, sizeof(um)); 13 for (int i = 0; i <= n; i++) 14 g[i].clear(); 15 } 16 void inserts(int u, int v) 17 { 18 g[u].push_back(v); 19 } 20 bool searchP() 21 { 22 queue<int> q; 23 dis = INF; 24 memset(dx, -1, sizeof(dx)); 25 memset(dy, -1, sizeof(dy)); 26 for (int i = 1; i <= n; i++) 27 if (um[i] == -1) 28 { 29 q.push(i); 30 dx[i] = 0; 31 } 32 while (!q.empty()) 33 34 { 35 int u = q.front(); 36 q.pop(); 37 if (dx[u] > dis) 38 break; 39 for (int i = 0; i < g[u].size(); i++) 40 { 41 int v = g[u][i]; 42 if (dy[v] == -1) 43 { 44 dy[v] = dx[u] + 1; 45 if (vm[v] == -1) 46 dis = dy[v]; 47 else 48 { 49 dx[vm[v]] = dy[v] + 1; 50 q.push(vm[v]); 51 } 52 } 53 } 54 } 55 return dis != INF; 56 } 57 bool dfs(int u) 58 { 59 for (int i = 0; i < g[u].size(); i++) 60 { 61 int v = g[u][i]; 62 if (!vis[v] && dy[v] == dx[u] + 1) 63 { 64 vis[v] = 1; 65 if (vm[v] != -1 && dy[v] == dis) 66 continue; 67 if (vm[v] == -1 || dfs(vm[v])) 68 { 69 vm[v] = u; 70 um[u] = v; 71 return 1; 72 } 73 } 74 } 75 return 0; 76 } 77 int maxMatch() 78 { 79 int res = 0; 80 while (searchP()) 81 { 82 memset(vis, 0, sizeof(vis)); 83 for (int i = 1; i <= n; i++) 84 if (um[i] == -1 && dfs(i)) 85 res++; 86 } 87 return res; 88 } 89 }MM;
1 const int N=1e6+10; 2 const int M=26; 3 int ch[N][M]; 4 int val[N]; 5 int sz; 6 void init() 7 { 8 sz = 1; 9 memset(ch[0], 0, sizeof(ch[0])); 10 } 11 inline int idx(char c) 12 { 13 return c - 'a'; 14 } 15 void insert(char *s,int v) 16 { 17 int u = 0, i; 18 for (i = 0; s[i]; i++) 19 { 20 int c = idx(s[i]); 21 if (!ch[u][c]) 22 { 23 memset(ch[sz], 0, sizeof(ch[sz])); 24 val[sz] = 0; 25 ch[u][c] = sz++; 26 } 27 u = ch[u][c]; 28 } 29 val[u] = v; 30 //val[u]=max(val[u],i);//val记录长度 31 } 32 int f[N],last[N];//last数组连接着所有相同后缀结束的模板串 33 void getFail() 34 { 35 queue<int> q; 36 f[0] = 0; 37 for (int c = 0; c < M; c++) 38 { 39 int u = ch[0][c]; 40 if (u) 41 { 42 f[u] = 0; 43 q.push(u); 44 last[u] = 0; 45 } 46 } 47 while (!q.empty()) 48 { 49 int r = q.front(); 50 q.pop(); 51 for (int c = 0; c < M; c++) 52 { 53 int u = ch[r][c]; 54 if (!u) 55 continue; 56 q.push(u); 57 58 int v = f[r]; 59 while (v && !ch[v][c]) 60 v = f[v]; 61 f[u] = ch[v][c]; 62 last[u] = val[f[u]] ? f[u] : last[f[u]]; 63 } 64 } 65 } 66 void print(int i,int j)//输出所有匹配位置 67 { 68 if (j) 69 { 70 //找到后的处理,i是结束位置,字符串长度或值为 val[i] 71 print(i, last[j]); 72 } 73 } 74 int find(char *T) 75 { 76 int j = 0, i = 0; 77 for (; T[i]; i++) 78 { 79 int c = idx(T[i]); 80 while (j && !ch[j][c]) 81 j = f[j]; 82 j = ch[j][c]; 83 if (val[j]) 84 print(i, j); 85 else if (last[j]) 86 print(i, last[j]); 87 } 88 return i; 89 }
1 const int maxc=2; 2 const int BIT=31; 3 int ch[N<<5][maxc],si[N<<5]; 4 void init() 5 { 6 //每次建一棵新的字典树都要初始化 7 memset(ch[0], 0, sizeof(ch[0])); 8 si[0] = 1; 9 } 10 void insert(int s) 11 { 12 int u = 0; 13 for (int i = BIT - 1; i >= 0; i--) 14 { 15 int c = (s & (1 << i) ? 1 : 0); 16 if (!ch[u][c]) 17 { 18 memset(ch[si[0]], 0, sizeof(ch[0])); 19 si[si[0]] = 0; 20 ch[u][c] = si[0]++; 21 } 22 u = ch[u][c]; 23 si[u]++; 24 } 25 } 26 int find(int s) 27 { 28 int u = 0; 29 for (int i = BIT - 1; i >= 0; i--) 30 { 31 int c = (s & (1 << i) ? 1 : 0); 32 if (!ch[u][c] || si[ch[u][c]] <= 0) 33 return 0; 34 u = ch[u][c]; 35 } 36 return 1; 37 } 38 bool erase(int s) 39 { 40 int u = 0; 41 for (int i = BIT - 1; i >= 0; i--) 42 { 43 int c = (s & (1 << i) ? 1 : 0); 44 if (!ch[u][c] || si[ch[u][c]] <= 0) 45 return 0; 46 u = ch[u][c]; 47 si[u]--; 48 } 49 return 1; 50 }
1 int ch[N*20][2],si[N*20]; 2 //1 bro 0 son 3 char val[N*20]; 4 void init() 5 { 6 ch[0][0] = ch[0][1] = 0; 7 si[0] = 1; 8 val[0] = 0; 9 } 10 void insert(char *s) 11 { 12 int u = 0; 13 for (int i = 0; s[i]; i++) 14 { 15 int v = ch[u][0]; 16 while (v && val[v] != s[i]) 17 v = ch[v][1]; 18 if (!v) 19 { 20 //链表头插 21 int &t = si[0]; 22 ch[t][1] = ch[u][0]; 23 24 si[t] = ch[t][0] = 0; 25 val[t] = s[i]; 26 v = ch[u][0] = t++; 27 } 28 u = v; 29 } 30 si[u] = 1; 31 } 32 bool find(char *s) 33 { 34 int u = 0; 35 for (int i = 0; s[i]; i++) 36 { 37 for (u = ch[u][0]; u && val[u] != s[i]; u = ch[u][1]) 38 ; 39 if (!u) 40 return 0; 41 } 42 if (si[u]) 43 return 1; 44 return 0; 45 } 46 //字符串哈希 H(i)=H(i+1)x+s[i],Hash(i,L)=H(i)-H(i+L)*x^L; 47 int bfind() 48 { 49 int l = 1, r = n; 50 while (l < r) 51 { 52 int mid = r - (r - l) / 2; 53 if (hash_l(mid)) 54 l = mid; 55 else 56 r = mid - 1; 57 } 58 return l; 59 }
1 const int N=1e6+10; 2 int bit=8; 3 int base=1<<bit; 4 int mask=base-1; 5 int maxbit=20; 6 int n,x[N],h,next[N],m; 7 int head[N],tial[N]; 8 void radixsort(int &h,int *x) 9 { 10 for (int i = 0; i < maxbit; i += bit) 11 { 12 memset(head, 0, sizeof(head[0]) * base); 13 for (int j = h, tmp; j; j = tmp) 14 //分桶 15 16 { 17 tmp = next[j]; 18 int c = (x[j] >> i) & mask; 19 if (!head[c]) 20 { 21 head[c] = tial[c] = j; 22 next[j] = 0; 23 } 24 else 25 { 26 next[tial[c]] = j; 27 next[j] = 0; 28 tial[c] = j; 29 } 30 } 31 h = 0; 32 int ti; 33 //for(int j=0;j<base;j++) if(head[j]) //合并,升序 34 for (int j = mask; j >= 0; j--) 35 if (head[j]) //合并，降序 36 { 37 if (!h) 38 h = head[j]; 39 else 40 next[ti] = head[j]; 41 ti = tial[j]; 42 } 43 } 44 } 45 void init() 46 { 47 bit = 0; 48 while (1 << bit < n) 49 bit++; 50 bit -= 3; 51 bit = std::min(20, bit); //最大的比 maxbit小一点 52 bit = std::max(4, bit); 53 base = 1 << bit; 54 mask = base - 1; 55 } 56 int main() 57 { 58 while (scanf("%d%d", &n, &m) == 2) 59 { 60 for (int i = 1; i <= n; i++) 61 { 62 scanf("%d", x + i); 63 x[i] += 500000; 64 next[i] = i + 1; 65 } 66 init(); 67 h = 1; 68 next[n] = 0; 69 radixsort(h, x); 70 71 for (int i = h; i && m; i = next[i]) 72 printf("%d%c", x[i] - 500000, (--m) ? ' ' : ' '); 73 return 0; 74 } 75 }
1 template <class T> 2 void qsort(int l,int r,T *p) 3 { 4 if (l >= r) 5 return; 6 T t = p[l]; 7 int i = l, j = r; 8 while (i < j) 9 { 10 //以第一个为基准排序，大的放左边，小的放右边 11 while (i < j && p[j]->score <= t->score) 12 j--; //找到一个小的 13 p[i] = p[j]; 14 while (i < j && p[i]->score > t->score) 15 i++; //找到一个大的 16 p[j] = p[i]; 17 } 18 p[i] = t; 19 qsort(l, i - 1, p); 20 qsort(i + 1, r, p); 21 }
1 template<class T> 2 void mergesort(int l,int r,T *p,T *tmp) 3 { 4 if (l == r) 5 return; 6 int mid = r - (r - l) / 2; 7 mergesort(l, mid - 1, p, tmp); 8 mergesort(mid, r, p, tmp); 9 int i = l, j = mid, k = l; 10 while (i < mid || j <= r) 11 { 12 //归并 13 if (j > r || i < mid && p[i]->score >= p[j]->score) 14 tmp[k++] = p[i++]; 15 else 16 tmp[k++] = p[j++]; 17 } 18 while (l <= r) 19 p[l] = tmp[l++]; 20 //复制回原数组 21 }
1 template <class T> 2 void HeapAdjust(T *p,int s,int m) 3 { 4 T rc = p[s]; 5 //沿 key较大的孩子结点向下筛选 6 for (int j = s << 1; j <= m; j <<= 1) 7 { 8 if (j < m && p[j] < p[j + 1]) 9 ++j; 10 if (rc >= p[j]) 11 break; 12 p[s] = p[j]; 13 s = j; 14 //j为 key较大的记录的下标 15 //rc应插入在位置 s上 16 } 17 p[s] = rc; 18 //插入 19 } 20 template <class T> 21 void HeapSort(T *p) 22 { 23 //数组下标从 1开始 24 for (int i = n / 2; i > 0; --i) 25 HeapAdjust(p, i, n); //建堆,不断将子二叉树调整为堆，然后整个二叉树就满足 26 了堆性质 27 for (int i = n, j = 1; j <= m; i--, j++) 28 //每次将当前最大(小)(堆顶)元素,放到尾部，调整堆 29 { 30 swap(p[1], p[i]); 31 HeapAdjust(p, 1, i - 1); 32 } 33 }
1 template <class T> 2 void HeapUpAdjust(T *p,int s)//从 s向上到 1调整堆 3 { 4 int x = p[s]; 5 while (s > 1) 6 { 7 if (x < p[s >> 1]) 8 p[s] = p[s >> 1]; 9 else 10 break; 11 s >>= 1; 12 } 13 p[s] = x; 14 } 15 template <class T> 16 void HeapDownAdjust(T *p,int s,int n)//从 s向下到 n调整堆 17 { 18 int x = p[s]; 19 s <<= 1; 20 while (s <= n) 21 { 22 if (s < n && p[s | 1] < p[s]) 23 s |= 1; 24 if (p[s] >= x) 25 break; 26 p[s >> 1] = p[s]; 27 28 s <<= 1; 29 p[s >> 1] = x; 30 } 31 }
优先队列入队：加元素到尾部，然后向上调整，size++
优先队列出队：将最后一个元素放到顶部，向下调整堆，size--
1 int n,a[N],d[N][20]; 2 void RMQ_init() 3 { 4 for (int i = 1; i <= n; i++) 5 d[i][0] = a[i]; 6 for (int j = 1; (1 << j) <= n; j++) 7 for (int i = 1; i + (1 << j) - 1 <= n; i++) 8 d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]); 9 } 10 int RMQ(int L,int R) 11 { 12 int k = 0; 13 while ((1 << (k + 1)) <= R - L + 1) 14 k++; 15 return min(d[L][k], d[R - (1 << k) + 1][k]); 16 }
公平游戏:一个游戏者可以把状态 A变成状态 B,另一个也可以.象棋是不公平游戏
规则 1:一个状态是必败状态当且仅当它的所有后继都是必胜状态
规则 2:一个状态是必胜状态当且仅当它至少有一个后继是必败状态
dp时,对于每个状态,查找它的后继状态有没有必败状态,有就是必胜状态,没有就是必败状态
当数据较大,dp打表会超时时就打出一个比较小的表找数(0出现)的规律
1 struct Point 2 { 3 double x, y; 4 Point(double x = 0, double y = 0) : x(x), y(y) {} 5 }; 6 typedef Point Vector; 7 Vector operator + (Vector A,Vector B) { 8 return Vector(A.x + B.x, A.y + B.y); } 9 Vector operator - (Point A,Point B) { 10 return Vector(A.x - B.x, A.y - B.y); } 11 Vector operator * (Vector A,double p) { 12 return Vector(A.x * p, A.y * p); } 13 Vector operator / (Vector A,double p) { 14 return Vector(A.x / p, A.y / p); } 15 bool operator < (const Point & a,const Point &b) { 16 return a.x < b.x || (a.x == b.x && a.y < b.y); } 17 const double eps=1e-9; 18 const double pi=acos(-1); 19 int dcmp(double x) { 20 if (fabs(x) < eps) 21 return 0; 22 else 23 return x < 0 ? -1 : 1; } 24 bool operator == (const Point& a,const Point &b) { 25 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; } 26 27 28 //基本运算 29 double Dot(Vector A,Vector B) //向量的点积 30 { 31 return A.x * B.x + A.y * B.y; 32 } 33 double Length(Vector A) //向量的长度 34 { 35 return sqrt(Dot(A, A)); 36 } 37 double Angle(Vector A,Vector B) //两个向量的夹角 38 { 39 return acos(Dot(A, B) / Length(A) / Length(B)); 40 } 41 double Cross(Vector A,Vector B) //向量叉积 42 { 43 return A.x * B.y - A.y * B.x; 44 } 45 double Area2(Point A,Point B,Point C)//三角形面积 46 { 47 return Cross(B - A, C - A); 48 } 49 Vector Rotate(Vector A,double rad)//向量逆时针旋转 50 { 51 return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); 52 } 53 Vector Normal(Vector A) 54 {//计算向量的单位法向量(左转 90度以后把长度归一化),调用前确保 A不是 0向量 55 double L = Length(A); 56 return Vector(-A.y / L, A.x / L); 57 } 58 //点和直线 59 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w) 60 {//调用前确保两条直线 P+tv和 Q+tw有唯一交点,即 Cross(v,w)非 0 61 Vector u = P - Q; 62 double t = Cross(w, u) / Cross(v, w); 63 return P + v * t; 64 } 65 double DistanceToLine(Point P,Point A,Point B) 66 {//点到直线的距离 67 Vector v1 = B - A, v2 = P - A; 68 return fabs(Cross(v1, v2)) / Length(v1); //不取绝对值得到有向距离 69 } 70 71 72 double DistanceToSegment(Point P,Point A,Point B) 73 {//点到线段的距离 74 if (A == B) 75 return Length(P - A); 76 Vector v1 = B - A, v2 = P - A, v3 = P - B; 77 if (dcmp(Dot(v1, v2)) < 0) 78 return Length(v2); 79 else if (dcmp(Dot(v1, v3)) > 0) 80 return Length(v3); 81 else 82 return fabs(Cross(v1, v2)) / Length(v1); 83 } 84 Point GetLineProjection(Point P,Point A,Point B) 85 {//点在直线上的投影 86 Vector v = B - A; 87 return A + v * (Dot(v, P - A) / Dot(v, v)); 88 } 89 bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2) 90 {//判线段相交,两线段恰好有一个公共点且不再任何一条线段的端点 91 double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1); 92 double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1); 93 return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; 94 } 95 bool PointOnSegment(Point p,Point a1,Point a2) 96 {//判断点是否在线段上(不含端点,端点直接判 p和 a1或 a2是否相等) 97 //if(p==a1||p==a2) return 1;//判点相同 98 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0; 99 } 100 //多边形 101 double PolygonArea(Point *p,int n) 102 {//多边形有向面积 103 double area = 0; 104 for (int i = 1; i < n - 1; i++) 105 area += Cross(p[i] - p[0], p[i + 1] - p[0]); 106 return area / 2; 107 } 108 int isPointInPolygon(Point p,Point*poly,int n) 109 {//点在多边形判定 110 int wn = 0; 111 for (int i = 0; i < n; i++) 112 { 113 if (PointOnSegment(p, poly[i], poly[(i + 1) % n])) 114 return -1; //在边界上 115 int k = dcmp(Cross(poly[(i + 1) % n] - poly[i], p - poly[i])); 116 int d1 = dcmp(poly[i].y - p.y); 117 int d2 = dcmp(poly[(i + 1) % n].y - p.y); 118 if (k > 0 && d1 <= 0 && d2 > 0) 119 wn++; 120 if (k < 0 && d2 <= 0 && d1 > 0) 121 wn--; 122 } 123 if (wn != 0) 124 return 1; //内部 125 return 0; 126 //外部 127 } 128 129 130 //计算凸包,输入点数组 p,个数为 p,输出点数组 ch.函数返回凸包顶点数 131 //输入不能有重复点.注意:函数执行完之后输入点的顺序被破坏 132 //如果不希望在凸包的边上有输入点,把两个<=改成< 133 //在精度要求高时建议用 dcmp比较 134 int ConvexHull(Point* p,int n,Point *ch) 135 { 136 sort(p, p + n); 137 int m = 0; 138 for (int i = 0; i < n; i++) 139 { 140 while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) 141 m--; 142 ch[m++] = p[i]; 143 } 144 int k = m; 145 for (int i = n - 2; i >= 0; i--) 146 { 147 while (m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) 148 m--; 149 ch[m++] = p[i]; 150 } 151 if (n > 1) 152 m--; 153 return m; //凸包顶点数 154 } 155 double PointoRecDist(Point p,Point ra,Point rb) 156 {//点到长方形的距离 157 if (ra.x > rb.x) 158 swap(ra.x, rb.x); //变成左上和右下角 159 if (ra.y > rb.y) 160 swap(ra.y, rb.y); 161 double dx = 0, dy = 0; 162 if (p.x < ra.x) 163 dx = ra.x - p.x; 164 else if (p.x > rb.x) 165 dx = p.x - rb.x; 166 if (p.y < ra.y) 167 dy = ra.y - p.y; 168 else if (p.y > rb.y) 169 dy = p.y - rb.y; 170 return sqrt(dx * dx + dy * dy); 171 } 172 double getcycy(double x,double neg,Point cyc,double ri) 173 {//由 x得到圆上点的 y坐标,neg表示上 1下-1 174 return cyc.y + neg * sqrt(ri * ri - (x - cyc.x) * (x - cyc.x)); 175 }
1 const double eps=1e-9; 2 int gauss(double a[N][N],bool *l,double *ans,int n) 3 { 4 int res = 0, r = 0; 5 for (int i = 0; i < n; i++) 6 l[i] = false; //全部初始化为自由元 7 for (int i = 0; i < n; i++) 8 { 9 for (int j = r; j < n; j++) 10 if (fabs(a[j][i] > eps)) 11 { 12 for (int k = i; k <= n; k++) 13 swap(a[j][k], a[r][k]); 14 break; 15 } 16 if (fabs(a[r][i]) < eps) 17 { 18 res++; 19 continue; 20 } 21 for (int j = 0; j < n; j++) 22 if (j != r && fabs(a[j][i]) > eps) 23 { 24 double tmp = a[j][i] / a[r][i]; 25 for (int k = i; k <= n; k++) 26 a[j][k] -= tmp * a[r][k]; 27 } 28 l[i] = true; 29 r++; 30 } 31 for (int i = 0; i < n; i++) 32 if (l[i]) 33 for (int j = 0; j < n; j++) 34 if (fabs(a[j][i]) > 0) 35 ans[i] = a[j][n] / a[j][i]; 36 return res; 37 }
1 //若只是记录某个数是否存在可用 bitset优化，原来已存在的数记为 bitset x ； 2 //则 x中所有标记位为 1的数+a[i]之后变成 x=x<<a[i]|x 3 //如果数最大小于 64可用 long long代替 bitset,更快,最大小于 128时用__int128 4 //无限背包有排列种数 5 //(1,4,9,...,17^2 中可重复选能有多少种方法拼成 1~300,114,141,411,111111 4种拼成 6的方法) 6 memset(dp,0,sizeof(dp)); 7 dp[0]=1; 8 for(int i=0;i<=300;i++) //每个数都可以往后面加 1,4,...,17^2 9 for(int j=1;j*j+i<=300;j++) 10 dp[j*j+i]+=dp[i];//取多少个 i 11 //无限背包组合种数 12 //(1,4,9,...,17^2 中可重复选能有多少种方法拼成 1~300,114,111111 2种拼成 6的方法) 13 memset(dp,0,sizeof(dp)); 14 dp[0]=1; 15 for(int i=1;i*i<=300;i++)//取的数递增,即取了 1后只能取 4再 9再...,不能继续取 1了 16 for(int j=0;j+i*i<=300;j++) 17 dp[j+i*i]+=dp[j]; 18 将 n拆分为 m个数(n个相同的球放入 m个盒子)的组合方法数<=>将 n拆分为最大数为 m相加的组合方法数 19 推论: 设 m<=n,n拆分成最多 m个数的和的拆分数等于将 n拆分成最大数不超过 m的拆分数
1 #include <bits/stdc++.h> 2 typedef long long LL; 3 const int N=1e6+10; 4 int cas=1,T; 5 int m,n; 6 char s[40][40]; 7 LL dp[2][N]; 8 const int HASH=30007; 9 int head[HASH],next[N],state[2][N],hn[2]; 10 inline int insert(int s,int *state,LL *dp) 11 { 12 int h = state[s] * 7 % HASH; //必要时乘个质数会散一些,如果还有一个数要变成(state[s]+x*1007)%HASH 13 int u = head[h]; 14 while (u) 15 { 16 if (state[u] == state[s]) 17 { 18 dp[u] += dp[s]; 19 return 0; 20 } 21 u = next[u]; 22 } 23 next[s] = head[h]; 24 head[h] = s; 25 return 1; 26 } 27 inline void ha(int x,int k,int d,int &h) 28 { 29 state[d ^ 1][h] = x; 30 dp[d ^ 1][h] = dp[d][k]; 31 h += insert(h, state[d ^ 1], dp[d ^ 1]); 32 } 33 LL ans[40][40]; 34 int main() 35 { 36 while (scanf("%d%d", &n, &m) == 2) 37 { 38 for (int i = 0; i < n; i++) 39 scanf("%s", s[i]); 40 for (int i = 0; i < n; i++) 41 for (int j = 0; j < m; j++) 42 s[i][j] = (s[i][j] == '.'); 43 int d = 0; 44 int M = m << 1; 45 memset(ans, 0, sizeof(ans)); 46 state[0][1] = 0; 47 dp[0][1] = 1; 48 hn[0] = 2; 49 for (int i = 0; i < n; i++) 50 { 51 52 for (int j = 0; j < m; j++, d ^= 1) 53 { 54 hn[d ^ 1] = 1; 55 if (!s[i][j]) //障碍 56 { 57 for (int k = 1, &h = hn[d ^ 1]; k < hn[d]; k++) 58 { 59 int t = state[d][k]; 60 if ((t & (3 << M)) || (t & 3)) 61 continue; 62 state[d ^ 1][h] = t << 2; 63 dp[d ^ 1][h] = dp[d][k]; 64 h++; 65 } 66 continue; 67 } 68 memset(head, 0, sizeof(head)); 69 for (int k = 1, &h = hn[d ^ 1]; k < hn[d]; k++) //每个非障碍格子都可以走 70 { 71 int t = state[d][k]; 72 if ((t & (3 << M)) && (t & 3)) 73 { 74 int x = t & 3; 75 int y = t & (3 << M); 76 t = t ^ x ^ y; 77 y >>= M; 78 if (x == 1 && y == 2) 79 { 80 if (t == 0) //没有没有其它插头时可以是一条(i,j)结束的回路 81 ans[i][j] += dp[d][k]; //() 82 } 83 else if (x == 2 && y == 1) 84 ha(t << 2, k, d, h); //)( 85 else if (x == 2 && y == 2) //)) 86 { 87 for (int l = 2, c = 0; l < M; l += 2) //往前找到匹配的(并改为) 88 { 89 if (((t >> l) & 3) == 2) 90 c--; 91 else if (((t >> l) & 3) == 1) 92 c++; 93 if (c == 1) 94 { 95 t = t ^ (3 << l); 96 break; 97 } 98 } 99 ha(t << 2, k, d, h); 100 } 101 else 102 { 103 //(( 104 for (int l = M - 2, c = 0; l > 0; l -= 2) //往后找到匹配的)并改为( 105 { 106 if (((t >> l) & 3) == 1) 107 c--; 108 109 else if (((t >> l) & 3) == 2) 110 c++; 111 if (c == 1) 112 { 113 t = t ^ (3 << l); 114 break; 115 } 116 } 117 ha(t << 2, k, d, h); 118 } 119 } 120 else if (!(t & (3 << M)) && !(t & 3)) //## 121 { 122 //if(s[i][j]) ha(t<<2,k,d,h);//不走 123 if (i + 1 < n && s[i + 1][j] && j + 1 < m && s[i][j + 1]) 124 ha(t << 2 | 6, k, d, h); 125 } 126 else 127 { 128 //#( #) (# )# 129 int x = t & 3; 130 int y = t & (3 << M); 131 if (x) 132 t ^= x; 133 else if (y) 134 t ^= y, x = y >> M; 135 if (i + 1 < n && s[i + 1][j]) 136 ha(t << 2 | (x << 2), k, d, h); 137 if (j + 1 < m && s[i][j + 1]) 138 ha(t << 2 | x, k, d, h); 139 } 140 } 141 } 142 } 143 for (int i = n - 1; i >= 0; i--) 144 { 145 for (int j = m - 1; j >= 0; j--) 146 if (s[i][j]) 147 { 148 printf("%lld ", ans[i][j]); 149 i = -1; 150 break; 151 } 152 } 153 } 154 return 0; 155 }
1 int n,a[N],idx[N]; 2 int lbig[N],lsmall[N],rsmall[N],rbig[N],bit[N],id[N]; 3 //lbig左边比当前数大,rbig右边比当前数大(不等于) 4 LL sum(int x) 5 { 6 LL rec = 0; 7 while (x) 8 { 9 rec += bit[x]; 10 x -= x & -x; 11 } 12 return rec; 13 } 14 LL add(int x) 15 { 16 while (x <= n) 17 { 18 bit[x]++; 19 x += x & -x; 20 } 21 } 22 void cal() 23 { 24 for (int i = 1; i <= n; i++) 25 idx[i] = a[i]; 26 sort(idx + 1, idx + n + 1); 27 memset(bit, 0, sizeof(bit)); 28 map<int, int> mp; //用于处理相同数 29 for (int i = 1; i <= n; i++) 30 { 31 id[i] = lower_bound(idx + 1, idx + 1 + n, a[i]) - idx; 32 lsmall[i] = sum(id[i] - 1); 33 add(id[i]); 34 int &x = mp[a[i]]; 35 lbig[i] = i - 1 - lsmall[i] - x; 36 x++; 37 } 38 memset(bit, 0, sizeof(bit)); 39 mp.clear(); 40 for (int i = n; i > 0; i--) 41 { 42 rsmall[i] = sum(id[i] - 1); 43 add(id[i]); 44 int &x = mp[a[i]]; 45 rbig[i] = n - i - rsmall[i] - x; 46 x++; 47 } 48 }
1 LL count(LL n,LL num) 2 { 3 LL e = 1, ans = 0; 4 for (int i = 0; n >= e; e *= 10, i++) 5 { 6 LL pre = n / e / 10; 7 LL dig = n / e % 10; 8 LL suf = n % e; 9 if (dig > num) 10 pre++; 11 if (!num) 12 pre--; //num==0时去掉前导 0的情况 13 ans += pre * e; 14 if (dig == num) 15 ans += suf; 16 } 17 if (n && !num) 18 ans++; //num==0且 n>0时 0也要算上 19 return ans; 20 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int N=1e5+10; 5 const int INF=0x3f3f3f3f; 6 const int bit=7; 7 int cas=1,T; 8 int dp[10][3]; 9 void init() 10 { 11 memset(dp, 0, sizeof(dp)); 12 dp[0][2] = 1; 13 for (int i = 1; i <= bit; i++) 14 { 15 dp[i][0] = dp[i - 1][0] * 10 + 2 * dp[i - 1][1] + dp[i - 1][2]; //62,4 16 dp[i][1] = dp[i - 1][1] + dp[i - 1][2]; //no62,4 head=2 17 dp[i][2] = 7 * dp[i - 1][1] + 8 * dp[i - 1][2]; //no62,4 head!=2 18 } 19 //for(int i=1;i<=bit;i++) printf("%d %d %d %d ",i,dp[i][0],dp[i][1],dp[i][2]); 20 } 21 int count(int n) 22 { 23 int ans = 0; 24 for (int i = bit, e = 1e6 + 0.5; i >= 0; i--, e /= 10) 25 { 26 if (e > n) 27 continue; 28 int pre = n / e % 10; //当前位数字 29 ans += pre * dp[i][0]; //0~pre-1 *e 30 if (pre > 6 || pre == 6 && n * 10 / e % 10 > 2) 31 ans += dp[i][1]; 32 //6+ no62,4 head=2 33 当前位 > 6 或者 当前位 = 6且下一位 > 2 34 35 if (pre > 4) ans += dp[i][2] + dp[i][1]; //4+ no62,4 head=2 + no62,4, head!=2 36 if (pre == 4 || n / e % 100 == 62) //+ 后边所有 37 { 38 ans += n % e; 39 break; 40 } 41 } 42 return ans; 43 } 44 int main() 45 { 46 init(); 47 int l, r; 48 //while(scanf("%d%d",&l,&r)==2 && (l&&r)) printf("%d %d ",count(r+1),count(l)); 49 while (scanf("%d%d", &l, &r) == 2 && (l && r)) 50 printf("%d ", r - l + 1 - count(r + 1) + count(l)); 51 return 0; 52 } 53 //数位 dp记忆化搜索,从高位往低位搜 54 //这里是[l,r]中数位中连续奇数数字有偶数个,连续偶数数字有奇数个 55 //第三维表示数位的数字,还没有数字为 2,奇数为 1,偶数为 0 56 //第四维表示数位的连续个数,还没有数字为 2,奇数个为 1,偶数个为 0 57 //第二维表示当前数位为 dig时有多少种 58 #include <bits/stdc++.h> 59 using namespace std; 60 typedef long long LL; 61 const int N=1e5+10; 62 const int INF=0x3f3f3f3f; 63 int cas=1,T; 64 LL dp[20][10][3][3];//dp[a][b][c][d]记忆化前面数位为 d个 c,当前位为 b,后面有 a个 9时的种数 65 LL dfs(int n,LL e,LL x,int odd,int even) 66 { 67 if (n < 0) 68 return odd != even; 69 if (e > x && odd == 2) 70 return dfs(n - 1, e / 10, x, odd, even); 71 int dig = x / e % 10; 72 if ((x + 1) % e == 0 && dp[n][dig][odd][even] != -1) 73 return dp[n][dig][odd][even]; //如果已经搜过直 74 接返回 75 if (odd == 2) 76 { 77 //还没有数位时不同转移 78 LL ans = 0; 79 ans += dfs(n - 1, e / 10, e - 1, 2, 2); //前导 0,还是没有数位 80 for (int i = 1; i < dig; i++) 81 ans += dfs(n - 1, e / 10, (i + 1) * e - 1, i & 1, 1); 82 ans += dfs(n - 1, e / 10, x % e, dig & 1, 1); 83 if ((x + 1) % e == 0) 84 dp[n][dig][odd][even] = ans; 85 return ans; 86 } 87 LL ans = 0; 88 for (int i = 0; i < dig; i++) 89 if ((i & 1) == odd || (i & 1) != odd && odd != even) 90 ans += dfs(n - 1, e / 10, (i + 1) * e - 1, i & 1, ((i & 1) == odd) ? (even ^ 1) : 1); 91 92 if ((dig & 1) == odd || (dig & 1) != odd && odd != even) 93 ans += dfs(n - 1, e / 10, x % e, dig & 1, ((dig & 1) == odd) ? (even ^ 1) : 1); 94 if ((x + 1) / 10 == e) 95 dp[n][dig][odd][even] = ans; //x=a99…99 时记忆化 96 return ans; 97 } 98 int main() 99 { 100 memset(dp, -1, sizeof(dp)); 101 scanf("%d", &T); 102 while (T--) 103 { 104 LL l, r; 105 scanf("%lld%lld", &l, &r); 106 LL lans = dfs(18, 1e18, l - 1, 2, 2); 107 //printf("lans:%lld ",lans); 108 LL rans = dfs(18, 1e18, r, 2, 2); 109 //printf("rans:%lld ",rans); 110 printf("Case #%d: %lld ", cas++, rans - lans); 111 } 112 return 0; 113 }
TSP问题:有 n个城市,两两之间均有道路相连,距离为 dis(i,j),求最短的路径经过每个点一次并回到原点
做法:状压 dp,O(n*2^n)
如果给定起点和终点 s,e,则可以用 O(n^2)dp 解决,每次在当前最小路径中插入下一个点获得加一个点进来后的最
小路径
//圆周涂色
把一个圆分成 n个扇形,每个扇形可以涂 m种颜色,要求相邻的颜色互不相同,有多少种涂法?
把 n个扇形的涂法记为 f(n),则 f(n)+f(n-1)=m*(m-1)^(n-1),包含第一个和第 n个扇形涂色相同的情况,相当于
将第 1个扇形和第 n个扇形合并成一个扇形的涂法,化简得 f(n)=(m-1)^n+(n&1?-1:1)*(m-1);
1 const int M=75; 2 int 3 ro[3][M]={{0,10,11,12,13,14,15,16,17,18,46,47,48,49,50,51,52,53,54,25,22,19,26,23,20,27,24, 4 21,9,8,7,6,5,4,3,2,1,39,42,45,38,41,44,37,40,43,36,35,34,33,32,31,30,29,28,//3*3*6=54个方格 5 63,59,55,62,64,56,58,66,65,60,57,61, 6 //十二条边 7 71,72,68,67,74,73,69,70},//8个顶点 8 {0,43,40,37,44,41,38,45,42,39,16,13,10,17,14,11,18,15,12,7,4,1,8,5,2,9,6,3,30,33,36,29,32,3 9 5,28,31,34,52,49,46,53,50,47,54,51,48,25,22,19,26,23,20,27,24,21, 10 62,58,61,66,55,57,65,63,59,56,60,64, 11 71,67,70,74,72,68,69,73}, 12 {0,7,4,1,8,5,2,9,6,3,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,4 13 2,43,44,45,10,11,12,13,14,15,16,17,18,48,51,54,47,50,53,46,49,52, 14 56,57,58,55,60,61,62,59,64,65,66,63, 15 68,69,70,67,72,73,74,71}}; 16 17 18 int pu[M],t[M],pemu[100][M],pn; 19 void count(int *v) 20 { 21 int vis[M]; 22 memset(vis, 0, sizeof(vis)); 23 for (int i = 1; i < M; i++) 24 if (!vis[i]) 25 { 26 int c = 0; 27 int u = i; 28 while (!vis[u]) 29 { 30 vis[u] = 1; 31 u = pu[u]; 32 c++; 33 } 34 v[c]++; 35 } 36 } 37 void init() 38 { 39 memset(pemu, 0, sizeof(pemu)); 40 pn = 0; 41 set<LL> s; 42 map<int, int> cc; 43 for (int i = 0; i < M; i++) 44 pu[i] = i; 45 for (int i = 1; i <= 4; i++) 46 { 47 for (int l = 1; l < M; l++) 48 t[l] = pu[ro[0][l]]; 49 memcpy(pu, t, sizeof(pu)); 50 for (int j = 1; j <= 4; j++) 51 { 52 for (int l = 1; l < M; l++) 53 t[l] = pu[ro[1][l]]; 54 memcpy(pu, t, sizeof(pu)); 55 for (int k = 1; k <= 4; k++) 56 { 57 for (int l = 1; l < M; l++) 58 t[l] = pu[ro[2][l]]; 59 memcpy(pu, t, sizeof(pu)); 60 //printf("i:%d j:%d k:%d ",i,j,k); 61 LL ha = 0; 62 for (int l = 1; l < M; l++) 63 ha = ha * 1000007 + pu[l]; 64 //for(int l=1;l<=12;l++) printf("%d ",pu[l]);printf(" "); 65 if (s.count(ha)) 66 continue; 67 s.insert(ha); 68 count(pemu[pn]); 69 int sum = 0; 70 for (int l = 1; l < M; l++) 71 printf("%d ", pemu[pn][l]), sum += pemu[pn][l]; 72 printf(" "); 73 cc[sum]++; 74 pn++; 75 } 76 } 77 } 78 for (map<int, int>::iterator it = cc.begin(); it != cc.end(); it++) 79 printf("%d %d ", it->first, it->second); 80 printf("%d ", s.size()); 81 }
1 //f(n)=a*x^n+b*x^n,得到 f(n)=c1*f(n-1)+c2*f(n-2),a*x^n=f(n)-b*x^n 2 //求 a^n%mod用矩阵快速幂求解 f(n)时 f(n)%mod的循环节为(mod*mod-1),最小循环节是(mod*mod-1)的约数 3 //即 f(n)%mod=f(n%(mod*mod-1))%mod 4 //当循环节为 p时 f(0)==f(p) && f(1)==f(p+1) 5 //可以暴力找循环节 6 const int M=2; 7 struct mar 8 { 9 LL a[M][M]; 10 mar operator*(const mar &b) 11 { 12 mar ans; 13 memset(&ans, 0, sizeof(ans)); 14 for (int i = 0; i < M; i++) 15 for (int j = 0; j < M; j++) 16 for (int k = 0; k < M; k++) 17 ans.a[i][j] = (ans.a[i][j] + a[i][k] * b.a[k][j]) % mod; 18 return ans; 19 } 20 }; 21 LL a[M]={2,1};//递推式 f(n)=a[0]f(n-1)+a[1]f(n-2)+...+a[M-1]f(n-M) 22 LL f[M]={1,0};//f[0]=f(M-1),f[1]=f(M-2),...,f(M-2)=f(1),f(M-1)=f(0) 倒过来的 23 LL mar_pow(LL n) 24 { 25 mar ans, t; 26 memset(&ans, 0, sizeof(ans)); 27 memset(&t, 0, sizeof(t)); 28 for (int i = 0; i < M; i++) 29 ans.a[i][i] = 1; //单位矩阵,相当于快速幂中初始化为 1 30 memcpy(t.a[0], a, sizeof(a)); 31 for (int i = 1; i < M; i++) 32 t.a[i][i - 1] = 1; 33 while (n) 34 { 35 if (n & 1) 36 ans = ans * t; 37 t = t * t; 38 n >>= 1; 39 } 40 LL res = 0; 41 for (int i = 0; i < M; i++) 42 res = (res + ans.a[M - 1][i] * f[i]) % mod; 43 //f(n)=f(n-1)*ans.a[M-1][i]+...+f(0)*ans.a[M-1][0]; 44 return res; 45 }
1 void dfs(int x) 2 { 3 int pos = upper_bound(b + 1, b + 1 + n, k / a[x]) - b - 1; 4 ans -= sum(pos); 5 for (int i = 0; i < u[x].size(); i++) 6 dfs(u[x][i]); 7 ans += sum(pos); 8 add(upper_bound(b + 1, b + 1 + n, a[x]) - b - 1); 9 } 10 struct node 11 { 12 int i, x, pos; 13 }; 14 void slove(int x)//由上面的 dfs改成的手写栈 15 { 16 stack<node> s; 17 int pos = upper_bound(b + 1, b + 1 + n, k / a[x]) - b - 1; 18 s.push(node{0, x, pos}); 19 while (!s.empty()) 20 { 21 node v = s.top(); 22 s.pop(); 23 if (v.i == u[v.x].size()) //dfs结束时 24 { 25 ans += sum(v.pos); 26 add(upper_bound(b + 1, b + 1 + n, a[v.x]) - b - 1); 27 continue; //return 28 } 29 x = u[v.x][v.i++]; //子树 30 s.push(v); 31 pos = upper_bound(b + 1, b + 1 + n, k / a[x]) - b - 1; 32 ans -= sum(pos); 33 s.push(node{0, x, pos}); //子树节点入栈 34 } 35 }
求<=n中除 p外所有质数的乘积%p
要解决这个问题首先要理解一个动态规划的定义，定义 S(v,p)为 1−v 中经过小于等于 p 的素数筛过之后的数的个数，
即此时整个数集里只包含<=p 的素数以及素因子>p 的数。
相应的有状态转移 S(v,p)=S(v,p−1)–(S(v/p,p−1)−S(p−1,p−1))
相应的可以定义状态 F(v,p)为 1−v 中经过小于等于 p 的素数筛过之后的数的乘积。相应的转移方程为：
F(v,p)=F(v,p−1)/(F(v/p,p−1)/F(p−1,p−1)∗pS(v/p,p−1)−S(p−1,p−1))
原题中要模的是 M，所以在定义 S(v,p)和相应的 F(v,p)时应当改为 1−v 中（除了 M,2M…）经过小于等于 p 的素数筛过
之后的数的个数，以及这些数的乘积，这个实际上影响到的是初始化的阶段。
实现上由于 v 的取值只需要用到 n/1,n/2…的值，所以分别用 l,h 的两种数组记录，其中 cl[x]表示的 S(x,p),ch[x]表示
S(n/x,p)，pl[x]表示 F(x,p)，ph[x]表示 F(n/x,p)。
1 syntax on 2 "高亮 3 set nu 4 "显示行号 5 set softtabstop=4 6 set shiftwidth=4 7 set smartindent 8 set scrolloff=7 9 set mouse=a 10 "按 tab缩进 4个空格大小 11 "默认缩进 4个空格大小（indent的缩进） 12 "自动缩进 13 "光标移动到 buffer的顶部和底部时保持 7行距离 14 "鼠标所有功能可用 15 set cursorline 16 "当前行下划线划出 17 "出现 swap文件时用这个 18 set nobackup 19 set noswapfile 20 winpos -4 -4 21 "设置窗口位置,使用 vim时自动最大化窗口 22 set lines=50 columns=200 "设定窗口大小 "按 F5编译运行 23 set guifont=UbuntuMono 14 24 colorscheme desert 25 "Ubuntu下 GVIM设置英文字体,字体名和字体大小 26 "F5编译运行 27 "w保存文本 ! g++ % -o %< 编译代码 ! ./%< linux执行 ! %<.exe windows执行 28 map <F5> :call Compile()<CR> 29 func Compile() 30 :w 31 :! g++ -std=c++11 % -o %< 32 :! ./%< 33 endfunc 34 "noremap insert模式（打字时的模式）下映射 35 "vmap/vnoremap visaul模式（选择时的模式）下映射 36 "map/nnoremap normal模式（按下 Esc后的模式）下映射 37 map <C-a> ggVG 38 vmap <C-c> "+y 39 map <C-v> "+p 40 map zz I//<Esc> 41 "Ctrl+a全选 42 "Ctrl+c选择后复制，可粘贴到外编辑器，如果只在 vim中粘贴按 y复制即可 43 "Ctrl+v normal模式下粘贴外部复制的文本 44 "zz行首注释 45 map tt I<Right><Right><Backspace><Backspace><Esc> "tt取消注释 46 "cpp文件自动输出，F2手动输出 47 auto BufNewFile *.cpp exec ":call Preset()" 48 map<F2> :call Preset()<CR> 49 func Preset() 50 "setline将当前行设为字符串，append在当前行后追加 51 call setline(line("."),"#include <bits/stdc++.h>") 52 let l=line(".")-1 53 let l=l+1 | call append(l,"using namespace std;") 54 let l=l+1 | call append(l,"typedef long long LL;") 55 let l=l+1 | call append(l,"const int N=1e5+10;") 56 let l=l+1 | call append(l,"const int INF=0x3f3f3f3f;") 57 58 59 let l=l+1 | call append(l,"const int mod=1e9+7;") 60 let l=l+1 | call append(l,"int cas=1,T;") 61 let l=l+1 | call append(l,"int main()") 62 let l=l+1 | call append(l,"{") 63 let l=l+1 | call append(l," //freopen("in","r",stdin);") 64 let l=l+1 | call append(l," //scanf("%d",&T);") 65 let l=l+1 | call append(l," return 0;") 66 let l=l+1 | call append(l,"}") 67 endfunc 68 """""""""""下面的不需要用到就不要打了 69 "F3输出头文件,bits/stdc++.h可用时不需要 70 map <F3> :call AddHeader()<CR> 71 func! AddHeader() 72 call setline(line("."),"#include <cstdio>") 73 let l=line(".")-1 74 let l=l+1 | call append(l,"#include <queue>") 75 let l=l+1 | call append(l,"#include <cstring>") 76 let l=l+1 | call append(l,"#include <string>") 77 let l=l+1 | call append(l,"#include <iostream>") 78 let l=l+1 | call append(l,"#include <cstdlib>") 79 let l=l+1 | call append(l,"#include <algorithm>") 80 let l=l+1 | call append(l,"#include <vector>") 81 let l=l+1 | call append(l,"#include <map>") 82 let l=l+1 | call append(l,"#include <set>") 83 let l=l+1 | call append(l,"#include <ctime>") 84 let l=l+1 | call append(l,"#include <cmath>") 85 let l=l+1 | call append(l,"#include <cctype>") 86 endfunc 87 "set guifont=Consolas:h14:cANSI 88 "windows下设置英文字体,字体名和字体大小
1 1）、单点增减+区间求和 2 思路：C[x]表示该点的元素:sum(x)=C[1]+C[2]+……C[x] 3 int arr[MAXN]; 4 inline int sum(int x){ 5 int res = 0; 6 while (x) 7 res += arr[x], x -= lowbit(x); 8 return res;} 9 inline void add(int x,int n){ 10 while (x < MAXN) 11 arr[x] += n, x += lowbit(x);} 12 inline int query(int x,int y){ 13 return sum(y) - sum(x - 1);} 14 （2）、区间增减+单点查询 15 思路：C[x]表示该点元素与左边元素的差值:num[x]=C[1]+C[2]+……C[x] 16 int arr[MAXN] 17 inline int sum(int x){ 18 int res = 0; 19 while (x) 20 res += arr[x], x -= lowbit(x); 21 return res;} 22 inline void add(int x,int n){ 23 while (x < MAXN) 24 arr[x] += n, x += lowbit(x);} 25 inline int update(int x,int y,int n){ 26 add(x, n); 27 add(y + 1, -n);} 28 （3）、区间增减+区间查询 29 思路：C1[x]表示该点元素与左边的差值，C2[x]表示的是 x*C[x] 30 sum(sum(C[j],j<=i)i<=x) 31 = x*C[1]+(x-1)*C[2]+……+C[x] 32 =(x+1)*sum(C[i],i<=x)-sum(i*C[i],i<=x); 33 则可以想到用 C1[x]维护 C[x]的值,C2[x]维护 x*C[X]的值 34 template <typename X> 35 struct tree_array{ 36 struct tree_array_single 37 { 38 X arr[MAXN]; 39 40 void add(int x, X n) 41 { 42 while (x <= N) 43 arr[x] += n, x += lowbit(x); 44 } 45 X sum(int x) 46 { 47 X sum = 0; 48 while (x) 49 sum += arr[x], x -= lowbit(x); 50 return sum; 51 } 52 } T1, T2; 53 void reset() 54 { 55 CLR(T1.arr, 0); 56 CLR(T2.arr, 0); 57 } 58 void add(int x, X n) 59 { 60 T1.add(x, n); 61 T2.add(x, x * n); 62 } 63 void update(int L, int R, int n) 64 { 65 add(L, n); 66 add(R + 1, -n); 67 } 68 X sum(int x) { return (x + 1) * T1.sum(x) - T2.sum(x); } 69 X query(int L, int R) { return sum(R) - sum(L - 1); } 70 };
1 ①单点增减（add） + 矩形求和（query） 2 ②矩形增减（update） + 单点求值（sum） 3 int arr[MAXN][MAXN] 4 inline void add(int x,int y,int n) { 5 for (int i = x; i < MAXN; i += lowbit(i)) 6 for (int j = y; j < MAXN; j += lowbit(j)) 7 arr[i][j] += n; 8 } 9 inline int sum(int x,int y){ 10 int res = 0; 11 for (int i = x; i; i -= lowbit(i)) 12 for (int j = y; j; j -= lowbit(j)) 13 res += arr[i][j]; 14 return res; 15 } 16 inline int query(int L,int B,int R,int T) { 17 return sum(R, T) + sum(L - 1, B - 1) - sum(R, B - 1) - sum(L - 1, T); 18 } 19 inline void update(int L,int B,int R,int T,int n){ 20 update(L, B, n); 21 update(L, T + 1, n); 22 update(R + 1, B, n); 23 update(R + 1, T + 1, n); 24 } 25 ③矩形增减（update）+ 矩形求和（query） 26 template<typename X> 27 class tree_array{ 28 struct tree_array_single 29 { 30 X arr[MAXN][MAXN]; 31 void add(int x, int y, X n) 32 { 33 for (int i = x; i < MAXN; i += lowbit(i)) 34 for (int j = y; j < MAXN; j += lowbit(j)) 35 arr[i][j] += n; 36 } 37 X sum(int x, int y) 38 { 39 X res = 0; 40 for (int i = x; i; i -= lowbit(i)) 41 for (int j = y; j; j -= lowbit(j)) 42 43 res += arr[i][j]; 44 return res; 45 } 46 } T1, T2, T3, T4; 47 void add(int x, int y, int n) 48 { 49 T1.add(x, y, n); 50 T2.add(x, y, y * n); 51 T3.add(x, y, x * n); 52 T4.add(x, y, x * y * n); 53 } 54 X sum(int x, int y) 55 { 56 return (x + 1) * (y + 1) * T1.sum(x, y) - (x + 1) * T2.sum(x, y) - (y + 1) * T3.sum(x, y) + T4.sum(x, y); 57 } 58 59 public: 60 void init() 61 { 62 CLR(T1.arr, 0); 63 CLR(T2.arr, 0); 64 CLR(T3.arr, 0); 65 CLR(T4.arr, 0); 66 } 67 void update(int L, int B, int R, int T, int n) 68 { 69 add(L, B, n); 70 add(L, T + 1, -n); 71 add(R + 1, B, -n); 72 add(R + 1, T + 1, n); 73 } 74 X query(int L, int B, int R, int T) 75 { 76 return sum(R, T) - sum(L - 1, T) - sum(R, B - 1) + sum(L - 1, B - 1); 77 } 78 }; 79 ④单点增减（add） + 立方体求和（query） 80 ⑤立方体增减（update） + 单点求值（sum） 81 int arr[MAXN][MAXN][MAXN]; 82 inline int sum(int x,int y,int z){ 83 int res = 0; 84 for (int i = x; i; i -= lowbit(i)) 85 for (int j = y; j; j -= lowbit(j)) 86 for (int k = z; k; k -= lowbit(k)) 87 res ^= arr[i][j][k]; 88 return res; 89 } 90 inline void add(int x,int y,int z,int n){ 91 for (int i = x; i < MAXN; i += lowbit(i)) 92 for (int j = y; j < MAXN; j += lowbit(j)) 93 for (int k = z; k < MAXN; k += lowbit(k)) 94 arr[i][j][k] += n; 95 } 96 inline void update(int x1,int y1,int z1,int x2,int y2,int z2,int n){ 97 add(x1, y1, z1, n); 98 add(x2 + 1, y1, z1, -n); 99 add(x1, y2 + 1, z1, -n); 100 add(x1, y1, z2 + 1, -n); 101 add(x2 + 1, y2 + 1, z1, n); 102 add(x2 + 1, y1, z2 + 1, n); 103 add(x1, y2 + 1, z2 + 1, n); 104 add(x2 + 1, y2 + 1, z2 + 1, -n); 105 } 106 inline int query(int x1,int y1,int z1,int x2,int y2,int z2){ 107 return sum(x2, y2, z2) - sum(x2, y2, z1 - 1) - sum(x2, y1 - 1, z2) - sum(x1 - 1, y2, z2) + sum(x2, y1 - 1, z1 - 1) + sum(x1 - 1, y2, z1 - 1) + sum(x1 - 1, y1 - 1, z2) - sum(x1 - 1, y1 - 1, z1 - 1); 108 } 109 110 111 ⑥立方体增减（update） + 立方体求和（query）///随便写写……复杂度较高 112 template<typename X> 113 class tree_array_Cube{ 114 struct tree_array_single 115 { 116 X arr[MAXN][MAXN][MAXN]; 117 X sum(int x, int y, int z) 118 { 119 X res = 0; 120 for (int i = x; i; i -= lowbit(i)) 121 for (int j = y; j; j -= lowbit(j)) 122 for (int k = z; k; k -= lowbit(k)) 123 res += arr[i][j][k]; 124 return res; 125 } 126 void add(int x, int y, int z, X n) 127 { 128 for (int i = x; i < MAXN; i += lowbit(i)) 129 for (int j = y; j < MAXN; j += lowbit(j)) 130 for (int k = z; k < MAXN; k += lowbit(k)) 131 arr[i][j][k] += n; 132 } 133 } T1, T2, T3, T4, T5, T6, T7, T8; 134 void add(int x, int y, int z, X n) 135 { 136 T1.add(x, y, z, n); 137 T2.add(x, y, z, x * n); 138 T3.add(x, y, z, y * n); 139 T4.add(x, y, z, z * n); 140 T5.add(x, y, z, x * y * n); 141 T6.add(x, y, z, y * z * n); 142 T7.add(x, y, z, x * z * n); 143 T8.add(x, y, z, x * y * z * n); 144 } 145 X sum(int x, int y, int z) 146 { 147 return (x + 1) * (y + 1) * (z + 1) * T1.sum(x, y, z) - (y + 1) * (z + 1) * T2.sum(x, y, z) - (x + 1) * (z + 1) * T3.sum(x, y, z) - (x + 1) * (y + 1) * T4.sum(x, y, z) + (z + 1) * T5.sum(x, y, z) + (x + 1) * T6.sum(x, y, z) + (y + 1) * T7.sum(x, y, z) - T8.sum(x, y, z); 148 } 149 150 public: 151 void init() 152 { 153 CLR(T1.arr, 0); 154 CLR(T2.arr, 0); 155 CLR(T3.arr, 0); 156 CLR(T4.arr, 0); 157 CLR(T5.arr, 0); 158 CLR(T6.arr, 0); 159 CLR(T7.arr, 0); 160 CLR(T8.arr, 0); 161 } 162 void update(int x1, int y1, int z1, int x2, int y2, int z2, X n) 163 { 164 add(x1, y1, z1, n); 165 add(x2 + 1, y1, z1, -n); 166 add(x1, y2 + 1, z1, -n); 167 add(x1, y1, z2 + 1, -n); 168 add(x2 + 1, y2 + 1, z1, n); 169 add(x2 + 1, y1, z2 + 1, n); 170 add(x1, y2 + 1, z2 + 1, n); 171 add(x2 + 1, y2 + 1, z2 + 1, -n); 172 } 173 X query(int x1, int y1, int z1, int x2, int y2, int z2) 174 { 175 return sum(x2, y2, z2) - sum(x2, y2, z1 - 1) - sum(x2, y1 - 1, z2) - sum(x1 - 1, y2, z2) + sum(x2, y1 - 1, z1 - 1) + sum(x1 - 1, y2, z1 - 1) + sum(x1 - 1, y1 - 1, z2) - sum(x1 - 1, y1 - 1, z1 - 1); 176 } 177 };
1 inline void init() 2 { 3 CLR(arr, 0); 4 for (int i = 1; i <= N; ++i) 5 for (int j = i; j <= N && arr[j] < num[i]; j += lowbit(j)) 6 arr[j] = num[i]; 7 } 8 inline int query(int L,int R) 9 { 10 int res = 0; 11 for (--L; L < R;) 12 { 13 if (R - lowbit(R) >= L) 14 { 15 res = max(res, arr[R]); 16 R -= lowbit(R); 17 } 18 else 19 { 20 res = max(res, num[R]); 21 --R; 22 } 23 } 24 return res; 25 } 26 inline void update(int x,int val) 27 { 28 int ori = num[x]; 29 num[x] = val; 30 if (val >= ori) 31 for (int i = x; i <= N && arr[i] < val; i += lowbit(i)) 32 arr[i] = val; 33 else 34 { 35 for (int i = x; i <= N && arr[i] == ori; i += lowbit(i)) 36 { 37 arr[i] = val; 38 for (int j = lowbit(i) >> 1; j; j >>= 1) 39 arr[i] = max(arr[i], arr[i - j]); 40 } 41 } 42 }