LightOJ1317---Throwing Balls into the Baskets 解题心得

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 Kturns.

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.

For each case, print the case number and the expected number of balls. Errors less than 10^-6 will be ignored.

2

1 1 1 0.5

1 1 2 0.5

Case 1: 0.5

Case 2: 1.000000

分析：

那个M完全没有用好吧，题目说这么长，完全就是在迷惑我们，结果就是  double ans = n*p*k;

代码：

#include <stdio.h>
#include <iostream> 
#include<cstdio>
#include<iomanip>
#include<cmath>
using namespace std;

int main()
{
    int t;
    int kase = 0;
    cin >> t;
    int n, m, k;
    double p;
    while (t--)
    {
        kase++;
        cin >> n >> m >> k >> p;
        double ans = n*p*k;
        cout << "Case " << kase << ‘:‘<<‘ ‘ << fixed << setprecision(6) << ans << endl;
    }
    return 0;
}