传送门
文章目录
- 题意：
- 思路：
题意：
给你nnn个灯，每次可以打开一个灯，当连续的kkk个灯有至少两个灯开着的时候停止，问最终期望能打开多少灯。
思路：
由于不想打latexlatexlatex，所以手推了公式。
实现起来就很简单啦。
ansansans初始为111是因为i=0i=0i=0的时候概率显然为111。
// Problem: E. Crypto Lights
// Contest: Codeforces - Deltix Round, Spring 2021 (open for everyone, rated, Div. 1 + Div. 2)
// URL: https://codeforces.com/problemset/problem/1523/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,k;
LL fun[N],inv[N];LL qmi(LL a,LL b) {LL ans=1;while(b) {if(b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;}return ans%mod;
}void init() {fun[0]=1;for(int i=1;i<N;i++) fun[i]=fun[i-1]*i%mod;inv[N-1]=qmi(fun[N-1],mod-2);for(int i=N-2;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
}LL C1(int n,int m) {return fun[n]*inv[n-m]%mod*inv[m]%mod;
}LL C2(int n,int m) {return fun[n-m]*fun[m]%mod*inv[n]%mod;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0); init();int _; scanf("%d",&_);while(_--) {scanf("%d%d",&n,&k);LL ans=1;for(int i=1;i<=n-1;i++) if(n-1ll*(i-1)*(k-1)>=i) ans+=C1(n-1ll*(i-1)*(k-1),i)*C2(n,i)%mod,ans%=mod;printf("%lld\n",ans%mod);}return 0;
}
/**/
