A. ><
大意: 给定一个序列相邻元素的大小关系, 求序列和的最小值
贪心, 最低点取0, 然后向左右延伸即可.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i) #define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int n, a[N], vis[N]; char s[N]; void solve(int x) { a[x] = 0; int t = x-1, now = 1; while (t>=0&&s[t+1]=='>') a[t--] = now++; t = x+1, now = 1; while (t<=n&&s[t]=='<') a[t++] = now++; } int main() { scanf("%s",s+1); n = strlen(s+1); if (s[1]=='<') solve(0); REP(i,1,n-1) if (s[i]=='>'&&s[i+1]=='<') solve(i); if (s[n]=='>') solve(n); REP(i,1,n) if (s[i]=='<'&&s[i+1]=='>') a[i] = max(a[i-1],a[i+1])+1; ll ans = 0; REP(i,0,n) ans += a[i]; printf("%lld ", ans); }
B. Two Contests
大意: 给定$n$个区间, 要求划分成两部分, 每部分的价值为区间交集的长度, 求最大价值.
假设某一部分交集为空, 那么另一部分只放一块最长的区间.否则的话, 两部分一定都有价值.
那么考虑右端点最小的一块区间$L$, 与$L$不相交的区间一定不与$L$在同一部分, 一定全在另一个区间内, 可以合并成一块区间$R$.
然后考虑剩余区间如何分配, 首先左端点一定在$L$的右端点左侧, 所以就可以按左端点从大到小排序, 枚举一下它们与$L$相交的长度即可.
最后要再特判一下$R$为空的情况
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i) #define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int n; pii a[N],b[N],c[N],L,R; pii add(pii a, pii b) { if (!a.x) return b; if (!b.x) return a; return pii(max(a.x,b.x),min(a.y,b.y)); } int len(pii a) { if (!a.x) return -INF; return a.y-a.x+1; } int main() { scanf("%d",&n); REP(i,1,n) scanf("%d%d",&a[i].x,&a[i].y); sort(a+1,a+1+n,[](pii a,pii b){return a.y-a.x<b.y-b.x;}); int ans = len(a[n]), cnt = 0; sort(a+1,a+1+n,[](pii a,pii b){return a.y<b.y;}); L = a[1]; REP(i,2,n) { pii t(add(L,a[i])); if (t.x>t.y) R = add(R,a[i]); else b[++cnt] = a[i]; if (R.x>R.y) return printf("%d ",ans),0; } sort(b+1,b+1+cnt,greater<pii>()); c[cnt] = b[cnt]; PER(i,1,cnt-1) c[i] = add(c[i+1],b[i]); ans = max(ans, len(add(c[1],L))+len(R)); if (!R.x) { int ma = -INF; REP(i,1,cnt) ma = max(ma, len(b[i])); ans = max(ans, len(add(c[1],L))+ma); } REP(i,1,cnt) { R = add(R,b[i]); ans = max(ans, len(add(c[i+1],L))+len(R)); } printf("%d ", ans); }
C. Neither AB nor BA
大意: 给定偶数$n$, 求所有长$n$的字符串个数, 满足只含$ABC$, 每次删除相邻两个字符, 不能删$AB$或$BA$, 最终可以删完.
假设奇数位$A_1$个$A$,$B_1$个$B$,偶数位$A_0$个$A$,$B_0$个$B$.
那么题目条件就等价于$A_1+B_0le frac{n}{2},A_0+B_1le frac{n}{2}$
用总方案减一下不合法方案即可.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i) #define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 998244353; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e7+10; int n, fac[N], ifac[N], po[N]; int C(int n, int m) { return (ll)fac[n]*ifac[m]%P*ifac[n-m]%P; } int main() { fac[0] = po[0] = 1; REP(i,1,N-1) fac[i]=(ll)fac[i-1]*i%P,po[i]=(ll)po[i-1]*2%P; ifac[N-1] = inv(fac[N-1]); PER(i,0,N-2) ifac[i]=(ll)ifac[i+1]*(i+1)%P; scanf("%d", &n); int ans = qpow(3,n); REP(i,n/2+1,n) ans = (ans-2ll*C(n,i)*po[n-i])%P; if (ans<0) ans += P; printf("%d ", ans); }
D.