Alice and Bob are playing a simple game. They line up a row of nn identical coins, all with the heads facing down onto the table and the tails upward.
For exactly mm times they select any kk 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 \le t \le 1000)t(1≤t≤1000) which is the total number of cases.
For each case, a line contains three space-separated integers nn, m (1 \le n, m \le 100)m(1≤n,m≤100) and k (1 \le k \le n)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.
样例输入复制
6 2 1 1 2 3 1 5 4 3 6 2 3 6 100 1 6 100 2
样例输出复制
0.500 1.250 3.479 3.000 5.500 5.000
题意：有n枚最初反面朝上的硬币，m次操作，每次选k个硬币重新抛掷，为了使硬币朝上的个数尽可能的多，问在最佳策略的情况下，硬币个数的期望。
dp[i][j]表示第i次操作后有j枚硬币朝上的概率，显然最优策略是挑选反面朝向的硬币重新抛掷，那么就会有两种情况
1：剩余反面朝上的硬币不少于k个，那么很简单，直接选k个反面朝上的硬币
2：剩余反面朝上的硬币少于k个，那么只能选所有反面朝上的硬币，剩余的用正面朝上的来补。
复杂度O(nmk)暴力所有情况。
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxm = 105;
double dp[maxm][maxm], p[maxm], C[maxm][maxm];
void init()
{
C[0][0] = 1;
for (int i = 1;i <= 100;i++)
{
C[i][0] = 1;
for (int j = 1;j <= i;j++)
C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
}
}
int main()
{
int n, i, j, k, sum, t, m, r;
p[0] = 1;init();
for (i = 1;i <= 100;i++)
p[i] = p[i - 1] / 2;
scanf("%d", &t);
while (t--)
{
scanf("%d%d%d", &n, &m, &k);
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for (i = 1;i <= m;i++)
{
for (j = 0;j <= n;j++)
{
for (r = 0;r <= k;r++)
{
if (n - j >= k)
dp[i][j + r] += dp[i - 1][j] * C[k][r] * p[k];
else dp[i][n - k + r] += dp[i - 1][j] * C[k][r] * p[k];
}
}
}
double ans = 0;
for (i = 1;i <= n;i++)
ans += dp[m][i] * i;
printf("%.3f\n", ans);
}
return 0;
}