ACM-ICPC 2017 Asia Urumqi A. Coins

Alice and Bob are playing a simple game. They line up a row of n identical coins, all with the heads facing down onto the table and the tails upward.

For exactly mm times they select any k of the coins and toss them into the air, replacing each of them either heads-up or heads-down with the same possibility. Their purpose is to gain as many coins heads-up as they can.

Input

The input has several test cases and the first line contains the integer t(1≤t≤1000) which is the total number of cases.

For each case, a line contains three space-separated integers n, m1≤n,m≤100) and k (1≤k≤n).

Output

For each test case, output the expected number of coins heads-up which you could have at the end under the optimal strategy, as a real number with the precision of 33 digits.

样例输入

6
2 1 1
2 3 1
5 4 3
6 2 3
6 100 1
6 100 2

样例输出

0.500
1.250
3.479
3.000
5.500
5.000

题目来源

ACM-ICPC 2017 Asia Urumqi

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long ll;
int t,n,m,k;
const int N=110;
double dp[N][N];
double    c[N][N];//ll也会爆,一定要写成double
double p[N];
double ans;
void C()
{
    c[0][0]=1;
    for(int i=1;i<=100;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(j==0||j==i) c[i][j]=1;
            else {
                c[i][j]=c[i-1][j-1]+c[i-1][j];
            }
        }
    }
    p[0]=1;
    for(int i=1;i<=100;i++) p[i]=p[i-1]/2;
}
int main()
{
    scanf("%d",&t);
    C();
    /*
    for(int i=1;i<=100;i++)
    {   printf("%dllll
",i);
        for(int j=0;j<=i;j++)
        {
            printf("%d%c",c[i][j],j==i?'
':' ');
        }
        //c[33][j]就爆int 了。
    }
    */
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        memset(dp,0,sizeof(dp));
        //dp[i][j]: i次操作后，正面朝上的面数为j的概率
        dp[0][0]=1;//初始都是反面
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<=n;j++)
            {
               for(int l=0;l<=k;l++)//取得k个里面l个正面朝上。
               {   //反面的个数大于k，直接从反面的里面取k个
                   if((n-j)>=k)  dp[i+1][j+l]+=dp[i][j]*c[k][l]*p[k];//取出的K个硬币每个的结果概率都是1/2（正/反）
                   //反面的个数小于k，只能再从已经是正面的里面再取出k-(n-j)个。
                   else  dp[i+1][j-(k-(n-j))+l]+=dp[i][j]*c[k][l]*p[k];
               }
            }
        }
        ans=0;
        for(int i=1;i<=n;i++)ans+=i*dp[m][i];//期望,别忘了*i
        printf("%.3f
",ans);
    }
    return  0;
}