lightOJ 1317 Throwing Balls into the Baskets

lightOJ  1317  Throwing Balls into the Baskets（期望）  解题报告

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88890#problem/A

题目：

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：

 #include<cstdio>
 #include<iostream>
 using namespace std;

 int t,n,m,k;
 double p,ans;
 int a[][];

 void init()
 {
     a[][]=;
     a[][]=;
     for(int i=;i<;i++)
     {
         a[i][i]=;
         a[i][]=;
         for(int j=;j<i;j++)
             a[i][j]=a[i-][j]+a[i-][j-];
     }
 }

 double count(int j)
 {
     double b=1.0;
     for(int i=;i<j;i++)
         b=b*p;//投中的期望
     for(int i=;i<n-j;i++)
         b=b*(1.0-p);//没投中的期望
     return b*j*a[n][j];
 }

 int main()
 {
     int c=;
     scanf("%d",&t);
     init();
     while(t--)
     {
         scanf("%d%d%d%lf",&n,&m,&k,&p);
         ans=0.0;//小数
         for(int i=;i<=n;i++)
             ans+=count(i);//第一轮的期望
         printf("Case %d: %.7lf\n",c++,ans*k);
     }
     return ;
 }