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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
#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\n",casex++,ans);
}
return 0;
}
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原文地址：http://blog.csdn.net/a1967919189/article/details/47840343
