A - A Count Task
找出所有的区间只含一种字母。
注意long long
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define pb push_back const int N=1e6+5; int main(){ int t; scanf("%d",&t); while(t--){ string s; cin>>s; ll len=s.size(); ll cur=0,ans=0,pos=0; for(ll i=0;i<len;i++){ if(s[i]==s[pos]){cur++;} else if(s[i]!=s[pos]){pos=i;ans+=cur*(cur+1)/2;cur=1;} } ans+=cur*(cur+1)/2; printf("%lld\n",ans); } // system("pause"); return 0; }View Code
C - A Path Plan
组合计数;
考虑所有情况,Cc(x1+y1,x1)*Cc(x2+y2,x2);
剔除相遇情况:Cc(x1+y2,x1)*Cc(x2+y1,x2);
#include<bits/stdc++.h> using namespace std; typedef long long ll; const ll mod=1e9+7; const int N=2e5+6; // const ll mod=998244353; ll fac[N],inv[N]; ll pow_mod(ll a,ll n) { ll ret =1; while(n) { if(n&1) ret=ret*a%mod; a=a*a%mod; n>>=1; } return ret; } void init() { fac[0]=1; for(int i=1;i<N;i++) { fac[i]=fac[i-1]*i%mod; } } ll Cc(ll x, ll y) { return fac[x]*pow_mod(fac[y]*fac[x-y]%mod,mod-2)%mod; } int main(){ init(); int t; scanf("%d",&t); while(t--){ int x1,x2,y1,y2; scanf("%d %d %d %d",&x1,&x2,&y1,&y2); ll a=(Cc(x1+y1,x1)*Cc(x2+y2,x2))%mod; ll b=(Cc(x1+y2,x1)*Cc(x2+y1,x2))%mod; ll ans=(a-b+mod)%mod; printf("%lld\n",ans); } // system("pause"); return 0; }View Code