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=;
 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[][]=;
     for(int i=;i<=;i++)
     {
         for(int j=;j<=i;j++)
         {
             if(j==||j==i) c[i][j]=;
             else {
                 c[i][j]=c[i-][j-]+c[i-][j];
             }
         }
     }
     p[]=;
     for(int i=;i<=;i++) p[i]=p[i-]/;
 }
 int main()
 {
     scanf("%d",&t);
     C();
     /*
     for(int i=1;i<=100;i++)
     {   printf("%dllll\n",i);
         for(int j=0;j<=i;j++)
         {
             printf("%d%c",c[i][j],j==i?'\n':' ');
         }
         //c[33][j]就爆int 了。
     }
     */
     while(t--)
     {
         scanf("%d%d%d",&n,&m,&k);
         memset(dp,,sizeof(dp));
         //dp[i][j]: i次操作后，正面朝上的面数为j的概率
         dp[][]=;//初始都是反面
         for(int i=;i<m;i++)
         {
             for(int j=;j<=n;j++)
             {
                for(int l=;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+][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+][j-(k-(n-j))+l]+=dp[i][j]*c[k][l]*p[k];
                }
             }
         }
         ans=;
         for(int i=;i<=n;i++)ans+=i*dp[m][i];//期望,别忘了*i
         printf("%.3f\n",ans);
     }
     return  ;
 }