Examining the Rooms HDU - 3625

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本文链接: https://blog.csdn.net/weixin_39132605/article/details/100590043

 https://cn.vjudge.net/problem/HDU-3625

总结:第一类斯特林数;

n个数组成k个循环排列;

s[i][j]表示i个数组成j个集合(k个无差别盒子):

单独一个组成一个集合s[i-1][j-1];插进一个集合中s[i-1][j]*(i-1) 有(i-1)种情况;

#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <set>
#include <map>
#include <stack>
//#include <tr1/unordered_map>
//#include <unordered_map>
#include <cmath>
//#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define sfl(i) scanf("%I64d",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%d\n",i)
#define prl(i) printf("%I64d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define inf 1e18
#define eps 1e-10
#define PI acos(-1.0)
#define lowbit(x) ((x)&(-x))
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define lson (rt<<1)
#define rson (rt<<1|1)
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

template<typename T>inline void read(T &x)
{
    x=0;
    static int p;p=1;
    static char c;c=getchar();
    while(!isdigit(c)){if(c=='-')p=-1;c=getchar();}
    while(isdigit(c)) {x=(x<<1)+(x<<3)+(c-48);c=getchar();}
    x*=p;
}

//-----------------------------------------------

const int maxn=1e2+9;
const int maxm=3e6+9;
const int mod=1e9+7;

ll s[maxn][maxn],f[maxn];

int main()
{
    //FAST_IO;
    //freopen("input.txt","r",stdin);

    s[1][0]=0;
    s[1][1]=1;
    for(int i=2;i<=20;i++)
    {
        for(int j=1;j<=i;j++)
        {
            s[i][j]=s[i-1][j]*(i-1)+s[i-1][j-1];
        }
    }

    f[1]=1;
    for(int i=2;i<=20;i++) f[i]=f[i-1]*i;


    int T;
    sfi(T);
    //cin>>T;
    while(T--)
    {
        int n,k;
        //cin>>n>>k;
        sfi(n);
        sfi(k);
        ll sum=0;
        for(int i=1;i<=k;i++)
        {
            sum+=s[n][i]-s[n-1][i-1];
        }
        printf("%.4lf\n",(double)sum/(double)f[n]);
    }

    return 0;
}

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转载自blog.csdn.net/weixin_39132605/article/details/100590043