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
程序分析：
首先我们要知道每一次扔的时候,0个人扔进的概率,1个人扔进的概率….
由于最后不关心M个篮子里球的具体情况,所以M其实没有什么用
只要区分扔不扔进去,至于扔到哪个篮子不管
dp[i][j]表示到第i轮,篮子里有j个球的概率
最后答案就是Σi?dp[k][i]
程序代码：
#include<iostream> #include<cstdio> using namespace std; int main() { int t,n,m,k,ans=1; double p; cin>>t; while(t--) { cin>>n>>m>>k>>p; printf("Case %d: %.6lf ",ans++, k*n*p); } return 0; }