ACM第六周竞赛题目——A LightOJ 1317

A - A

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 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

解题思路：

•题意：

•N个人， M个篮框，每个人投进球的概率是P。

•问每个人投K次， 总进球数的期望。

即 1 * C(1， 1) * 0.5^1 * 0.5 ^ 0

•每个人投球是相互独立的， 互不影响。求出一个人投K次的期望再乘以N即可。

•进球概率是P， 没有指定进哪一个篮框， 所以总的进球概率还是P（M* 1/M *P）

•每个人的进球数期望是：

•E = ∑（i * C(K, i) * P^i*(1 - P)^(K - i)）

•最终答案 = E * N.

程序代码：

#include<cstdio>
using namespace std;
int main()
{
    int t,Case=0;
    scanf("%d",&t);
    while(t--)
    {
     double p,ans;
     int m,n,k;
     scanf("%d%d%d%lf",&n,&m,&k,&p);
     ans=n*k*p;
    printf("Case %d: %.6lf
",++Case,ans);
    }
return 0;
}