light oj 1317

Description

You probably have played the game "Throwing Balls into the Basket". It is a simple game. You have to throw a ball into a basket from a certain distance. One day we (the AIUB ACMMER) were playing the game. But it was slightly different from the main game. In our game we were N people trying to throw balls into M identical Baskets. At each turn we all were selecting a basket and trying to throw a ball into it. After the game we saw exactly S balls were successful. Now you will be given the value of N and M. For each player probability of throwing a ball into any basket successfully is P. Assume that there are infinitely many balls and the probability of choosing a basket by any player is 1/M. If multiple people choose a common basket and throw their ball, you can assume that their balls will not conflict, and the probability remains same for getting inside a basket. You have to find the expected number of balls entered into the baskets after K turns.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing three integers N (1 ≤ N ≤ 16), M (1 ≤ M ≤ 100) and K (0 ≤ K ≤ 100) and a real number P (0 ≤ P ≤ 1). P contains at most three places after the decimal point.

Output

For each case, print the case number and the expected number of balls. Errors less than 10^-6 will be ignored.

Sample Input

2

1 1 1 0.5

1 1 2 0.5

Sample Output

Case 1: 0.5

Case 2: 1.000000

题意：

n个人 m个篮子 每一轮每一个人能够选m个篮子中一个扔球 扔中的概率都是p 求k轮后全部篮子里面球数量的期望值

思路：

依据期望 的定义与篮筐个数无关。由于题目如果互不影响。结果就是N*K*P。

代码：

#include<cstdio>
using namespace std;
int main()
{
    int T;
    int casex=1;
    double N,M,K,P;
    double ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf%lf%lf",&N,&M,&K,&P);
        ans=N*K*P;
        printf("Case %d: %lf
",casex++,ans);
    }
    return 0;
}