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
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 #include <vector>
6 #include <queue>
7 #include <set>
8 #include <map>
9 #include <string>
10 #include <cmath>
11 #include <cstdlib>
12 #include <ctime>
13 using namespace std;
14 typedef long long ll;
15 int t,n,m,k;
16 const int N=110;
17 double dp[N][N];
18 double c[N][N];//ll也会爆
19 double p[N];
20 double ans;
21 void C()
22 {
23 c[0][0]=1;
24 for(int i=1;i<=100;i++)
25 {
26 for(int j=0;j<=i;j++)
27 {
28 if(j==0||j==i) c[i][j]=1;
29 else {
30 c[i][j]=c[i-1][j-1]+c[i-1][j];
31 }
32 }
33 }
34 p[0]=1;
35 for(int i=1;i<=100;i++) p[i]=p[i-1]/2;
36 }
37 int main()
38 {
39 scanf("%d",&t);
40 C();
41 /*
42 for(int i=1;i<=100;i++)
43 { printf("%dllll\n",i);
44 for(int j=0;j<=i;j++)
45 {
46 printf("%d%c",c[i][j],j==i?‘\n‘:‘ ‘);
47 }
48 //c[33][j]就爆int 了。
49 }
50 */
51 while(t--)
52 {
53 scanf("%d%d%d",&n,&m,&k);
54 memset(dp,0,sizeof(dp));
55 //dp[i][j]: i次操作后，正面朝上的面数为j的概率
56 dp[0][0]=1;//初始都是反面
57 for(int i=0;i<m;i++)
58 {
59 for(int j=0;j<=n;j++)
60 {
61 for(int l=0;l<=k;l++)//取得k个里面l个正面朝上。
62 { //反面的个数大于k，直接从反面的里面取k个
63 if((n-j)>=k) dp[i+1][j+l]+=dp[i][j]*c[k][l]*p[k];
64 //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。
65 else dp[i+1][j-(k-(n-j))+l]+=dp[i][j]*c[k][l]*p[k];
66 }
67 }
68 }
69 ans=0;
70 for(int i=1;i<=n;i++)ans+=i*dp[m][i];//期望
71 printf("%.3f\n",ans);
72 }
73 return 0;
74 }
原文地址：https://www.cnblogs.com/tingtin/p/9397610.html