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 Npeople 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). Pcontains 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
题目大意：
有n个人，m个篮筐，一共打了k轮，每轮每个人可以投一个球，每个球投进的概率都是p,求k轮后，投中的球的期望是多少？
分析：
因为每个人投进的概率都是相同的，所以期望也是相同的。因此只需要求出第一轮的期望就可以了，总期望=k*第一轮的期望。
代码如下：
#include <stdio.h> int main() { int T,cas=0; scanf("%d",&T); while(T--) { cas++; double N,M,K; double P,ans; scanf("%lf%lf%lf%lf",&N,&M,&K,&P); ans=N*P*K; printf("Case %d: %lf ",cas,ans); } return 0; }