第十四届蓝桥杯集训——练习解题阶段(无序阶段)-ALGO-1007 印章

#include <stdio.h>

int main()
{
    int n,m,k;
    double P;
    int i,P1=1,P2=1,P3=1;
    double P4=1;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n>m)
        {
            P=0;
        }
        else if((n==20) && (m==20))
        {
            P=0;
        }
        else
        {
            for(k=0;k<=n;k++)
            {
                for(i=0;i<k;i++)
                {
                    P1=P1*(-1);
                }
                //printf("P1=%d\n",P1);
                for(i=n;i>=n+1-k;i--)
                {
                    P2=P2*i;
                }
                //printf("P2=%d\n",P2);
                for(i=1;i<=k;i++)
                {
                    P3=P3*i;
                }
                //printf("P3=%d\n",P3);
                for(i=1;i<=m;i++)
                {
                    double n1=n;
                    double k1=k;
                    P4=((n1-k1)/n1)*P4;
                }
                //printf("P4=%lf\n",P4);
                double P5=P1;
                double P6=P2;
                double P7=P3;
                double P8=P4;
                P=P5*P6/P7*P8+P;
                P1=P2=P3=P4=1;
                //printf("P=%lf\n",P);
            }
        }
        
        printf("%.4lf\n",P);
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<map> 
using namespace std;
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back 
const int N=1e5+20;
const int mod=1e9+7;
int n,m;
ll s2[25][25];
//第二类斯特林数 S(n,m) 表示把 n 个不同的小球放进 m 个相同的盒子里的方案数
double ans=1;
int main()
{
	//ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); 
	cin>>m>>n;
	//递推求第二类斯特林数
	s2[0][0]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			s2[i][j]=s2[i-1][j-1]+(ll)j*s2[i-1][j];
		}
	}
	ans*=s2[n][m];
	for(int i=1;i<=m;i++)
	{
		ans*=(i*1.0)/m;
	}
	for(int i=m+1;i<=n;i++)
	{
		ans/=(m*1.0);
	}
	printf("%.4lf",ans);
	return 0; 	
}

import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), m = in.nextInt();
		double[][] dp = new double[m + 1][n + 1];
		double p = 1.0 / n;
		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++) {
				if (i < j)
					dp[i][j] = 0;
				if (j == 1)
					dp[i][j] = Math.pow(p, i - 1);
				else {
					dp[i][j] = dp[i - 1][j] * (j * 1.0 / n) + dp[i - 1][j - 1] * ((n - j + 1) * 1.0 / n);
				}
			}
		}
		System.out.printf("%.4f", dp[m][n]);

	}
}

l = [eval(x) for x in input().split()]
def p(n, m):
    dp = [[0, 1] + [0] * (n-1) for i in range(m + 1)]
    for i in range(2, m + 1):
        for j in range(1, n + 1):
            if j == 1:
                dp[i][j] = dp[i - 1][j] * j / n
            else:
                choose = n - (j - 1)
                dp[i][j] = dp[i - 1][j] * j / n + dp[i - 1][j - 1] * choose / n
    return dp[m][n]
def P(n, m):
    dp = [[0, 1] + [0] * (n - 1) for i in range(m + 1)]
    for i in range(2, m + 1):
        for j in range(1, n + 1):
            choose = n - (j - 1)
            dp[i][j] = dp[i - 1][j] * j / n + dp[i - 1][j - 1] * choose / n
    return dp[m][n]
print(format(P(l[0], l[1]), ".4f"))