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
Output for Sample Input
2
1 1 1 0.5
1 1 2 0.5
Case 1: 0.5
Case 2: 1.000000
Problem Setter: Muhammad Rifayat Samee
Special Thanks: Jane Alam Jan
题意：n个人，m个篮筐，k轮投球，每个人一次投球进的概率为p，每一轮每个人都会投一次球，等概率选择每个篮筐而且之间互不影响。问k轮后进球数量的期望。
思路：
因为选择哪一个篮筐没有影响，所以直接忽略，投一次进的概率为p，一共会投n*k次，所以期望E = n*k * p。
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#define nmax 510
using namespace std;
int cas=0,T,m,k,n;
double p;
int main(){scanf("%d",&T); while(T--){scanf("%d%d%d%lf",&n,&m,&k,&p);printf("Case %d: %lf\n",++cas,n*k*p);}}