Problem 1002
因为是完全K叉树
所以这棵树的所有孩子最多只有一个不是满K叉树
对这个孩子进行递归处理即可
注意K=1时要特判
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int inf=(1<<30)-1; 4 const int maxn=100010; 5 #define REP(i,n) for(int i=(0);i<(n);i++) 6 #define FOR(i,j,n) for(int i=(j);i<=(n);i++) 7 typedef long long ll; 8 typedef pair<int,int> PII; 9 int IN(){ 10 int c,f,x; 11 while (!isdigit(c=getchar())&&c!='-');c=='-'?(f=1,x=0):(f=0,x=c-'0'); 12 while (isdigit(c=getchar())) x=(x<<1)+(x<<3)+c-'0';return !f?x:-x; 13 } 14 #define de(x) cout << #x << "=" << x << endl 15 #define MP make_pair 16 #define PB push_back 17 #define fi first 18 #define se second 19 ll n,k; 20 ll sum[100],t[100]; 21 22 int main() 23 { 24 int T;scanf("%d",&T); 25 while(T--) 26 { 27 scanf("%lld%lld",&n,&k); 28 if(k==1) { 29 if(n%4==0) printf("%lld ",n); 30 else if(n%4==1) printf("1 "); 31 else if(n%4==2) printf("%lld ",n+1); 32 else if(n%4==3) puts("0"); 33 continue; 34 } 35 if(k==2) { 36 37 } 38 ll m=1,res=0; 39 int tt=0; 40 for(tt=0;res<n;tt++) 41 { 42 sum[tt]=m;res+=m; 43 if(m*k+res<=n) m=m*k; 44 else m=n-res; 45 if(!tt) t[tt]=1; 46 else t[tt]=t[tt-1]*k+1; 47 } 48 ll ans=0; 49 if(sum[tt-1]&1) ans=1; 50 --tt; 51 ll M; 52 ll s1=k+1,s2=1,s3=0; 53 if(sum[tt]%k!=0) { 54 M=sum[tt]/k+1; 55 s3=sum[tt]%k+1; 56 } 57 else { 58 M=sum[tt]/k; 59 s3=k+1; 60 } 61 int len=2; 62 for(int i=tt-1;i>=0;i--) 63 { 64 if(M>=2&&((M-1)&1)) ans^=s1; 65 ans^=s3; 66 if((sum[i]-M)&1) ans^=s2; 67 ll j; 68 if(M%k==0) j=k;else j=M%k; 69 s3=s1*(j-1)+s3+(k-j)*s2+1; 70 if(M%k==0) M=M/k; 71 else M=M/k+1; 72 s1=t[len]; 73 s2=t[len-1]; 74 len++; 75 } 76 printf("%lld ",ans); 77 } 78 return 0; 79 }
Problem 1005
答案就是(n+1)/2+1
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int inf=(1<<30)-1; 4 const int maxn=100010; 5 #define REP(i,n) for(int i=(0);i<(n);i++) 6 #define FOR(i,j,n) for(int i=(j);i<=(n);i++) 7 typedef long long ll; 8 typedef pair<int,int> PII; 9 int IN(){ 10 int c,f,x; 11 while (!isdigit(c=getchar())&&c!='-');c=='-'?(f=1,x=0):(f=0,x=c-'0'); 12 while (isdigit(c=getchar())) x=(x<<1)+(x<<3)+c-'0';return !f?x:-x; 13 } 14 #define de(x) cout << #x << "=" << x << endl 15 #define MP make_pair 16 #define PB push_back 17 #define fi first 18 #define se second 19 int a,b,T; 20 int vis[maxn]; 21 int main() 22 { 23 scanf("%d",&T); 24 while(T--) 25 { 26 scanf("%d",&a); 27 printf("%d ",(a+1)/2+1); 28 } 29 return 0; 30 }
problem 1006
考虑分组背包
首先把数分组
2, 4, 6, 8, 10, 12, 14, 16, ...
3, 6, 9, 12, 15, 18, 21, 24...
5, 10, 15, 20, 25, 30...
7, 14, 21, 28, 35, 42...
11, 22, 33, 44, 55, 66...
13, 26, 39, 52, 65, 78...
17, 34, 51, 68, 85, 102...
19, 38, 57, 76, 95, 114...
23, 46, 69, 92...
29, 58, 87, 116..
31, 62, 93, 124...
...
...
这么分组之后我们发现，有某些数被分到了多组
比如12，因为12既可以被2整除也可以被3整除
仔细找规律我们发现从23开始的每个组的每个数都只属于一组。
而前8组可能会有重复的数的出现。
这样的话，我们把前面8组先预处理一下，
设f[mask][i]为mask状态下选了i个数的方案数。
然后再进行分组背包。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
typedef long long LL;
const int N = 510;
const LL mod = 1e9 + 7;
LL f[305][N];
int b1[N], b2[N], prime[N], state[N];
int cnt = 0;
int n, K;
int T;
LL ans;
void init(){
rep(i, 2, 503){
bool fl = true;
rep(j, 2, (int)sqrt(i + 0.5)) if (i % j == 0){
fl = false;
break;
}
if (fl) prime[cnt++] = i;
}
for (int i = 2; i * i <= 500; ++i)
for (int j = i * i; j <= 500; j += i * i) b1[j] = 1;
rep(i, 8, cnt - 1)
for (int j = 1; j * prime[i] <= 500; ++j) b2[j * prime[i]] = 1;
rep(i, 2, 500){
if (b1[i]) continue;
rep(j, 0, 7) if (i % prime[j] == 0) state[i] |= (1 << j);
}
}
int main(){
init();
scanf("%d", &T);
while (T--){
scanf("%d%d", &n, &K);
memset(f, 0, sizeof f);
f[0][0] = 1;
rep(i, 1, n){
if (b1[i] || b2[i]) continue;
dec(j, 255, 0) if (!(state[i] & j))
dec(k, K - 1, 0) (f[state[i] | j][k + 1] += f[j][k]) %= mod;
}
rep(i, 8, cnt - 1){
dec(k, K - 1, 0){
dec(j, 255, 0){
if (!f[j][k]) continue;
for (int times = 1, num; (num = times * prime[i]) <= n; ++times){
if (b1[num]) continue;
if (j & state[num]) continue;
(f[state[num] | j][k + 1] += f[j][k]) %= mod;
}
}
}
}
ans = 0;
rep(k, 1, K) rep(i, 0, 255) (ans += f[i][k]) %= mod;
printf("%lld
", ans);
}
return 0;
}
Problem 1008
考虑直线近似经过每一个点
每个点同时考虑2种情况
即不到一点和超过一点
极角排序后扫一圈
二分判断下另一边过哪里
维护前缀和后进行计算即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 typedef long long ll; 14 struct point 15 { 16 ll x,y,val; 17 point(){} 18 point(ll _x,ll _y):x(_x),y(_y) 19 {} 20 point operator -(const point &b) const 21 { 22 return point(x-b.x,y-b.y); 23 } 24 ll operator *(const point &b) const 25 { 26 return x*b.x+y*b.y; 27 } 28 ll operator ^(const point &b) const 29 { 30 return x*b.y-y*b.x; 31 } 32 }; 33 ll _,n; 34 point p[50010]; 35 ll sum[50010],pre[50010]; 36 ll cross(point sp,point ep,point op) 37 { 38 return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x); 39 } 40 inline bool cmp(const point &p1,const point &p2) 41 { 42 point o; 43 o.x=o.y=0; 44 return cross(p1,p2,o)>0; 45 } 46 ll calc(ll pos,ll l,ll r) 47 { 48 if (l>r) return 0; 49 l=l+pos; 50 r=r+pos; 51 if (l>n) l-=n; 52 if (r>n) r-=n; 53 if (l<=0) l+=n; 54 if (r<=0) r+=n; 55 if (l>r) 56 return (pre[n]-pre[l-1]+pre[r]); 57 else return pre[r]-pre[l-1]; 58 } 59 int main() 60 { 61 scanf("%lld",&_); 62 while (_--) 63 { 64 scanf("%lld",&n); 65 ll i; 66 for (i=1;i<=n;i++) 67 scanf("%lld%lld%lld",&p[i].x,&p[i].y,&p[i].val); 68 point o; 69 o.x=0; 70 o.y=0; 71 sort(p+1,p+n+1,cmp); 72 memset(pre,0,sizeof(pre)); 73 for (i=1;i<=n;i++) 74 pre[i]=pre[i-1]+p[i].val; 75 ll ans=0; 76 //cout<<p[1].x<<" "<<p[1].y<<endl; 77 //cout<<p[2].x<<" "<<p[2].y<<endl; 78 for (i=1;i<=n;i++) 79 { 80 ll l=1,r=n-1; 81 while (l<=r) 82 { 83 ll mid=(l+r)>>1; 84 ll pos=i+mid; 85 pos%=n; 86 //cout<<i<<" "<<mid<<" "<<pos<<endl; 87 ll x=cross(p[i],p[pos],o); 88 if (x<0) r=mid-1; 89 else l=mid+1; 90 } 91 ll up=calc(i,1,r); 92 ll down=calc(i,r+1,n-1); 93 ll sum=up*down; 94 //cout<<i<<" "<<r<<" "<<up<<" "<<down<<endl; 95 ans=max(ans,sum+p[i].val*up); 96 ans=max(ans,sum+p[i].val*down); 97 } 98 printf("%lld ",ans); 99 } 100 return 0; 101 }
Problem 1011
直接在数列的后面加数即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 int p[10000010]; 14 void init() 15 { 16 p[1]=1; 17 p[2]=2; 18 int j=2,i=2,x=2; 19 while (i<=10000000) 20 { 21 for (int k=1;k<=p[j];k++) 22 p[i++]=x; 23 j++; 24 if (x==1) x=2; else x=1; 25 } 26 } 27 int main() 28 { 29 init(); 30 int _; 31 scanf("%d",&_); 32 while (_--) 33 { 34 int x; 35 scanf("%d",&x); 36 printf("%d ",p[x]); 37 } 38 return 0; 39 }