LightOJ 1317 - Throwing Balls into the Baskets

最新推荐文章于 2020-04-23 15:06:34 发布

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最新推荐文章于 2020-04-23 15:06:34 发布

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文章标签：期望 LightOJ

本文链接：https://blog.youkuaiyun.com/abclydabclyd/article/details/76450851

本文介绍了一款名为“投球入筐”的游戏，并详细解析了如何计算多人多轮投掷后球成功进入筐中的期望值。文章通过数学方法简化了问题，提供了一个高效的解决方案。

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1317 - Throwing Balls into the Baskets

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

1 1 1 0.5

1 1 2 0.5

Case 1: 0.5

Case 2: 1.000000

PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE

SPECIAL THANKS: JANE ALAM JAN

题目大意：每一轮中，有n个人各执一球往m个篮子里扔球，各自的扔中概率皆为p，问k轮之后中球的期望。

思路：每一轮里每个人中秋的期望为p，n个人就为n*p,k轮之后的总期望为k*(n*p);

ac代码：

#include<stdio.h>

int main() {
	int t,n,m,k,cas=0;
	double p;

	scanf("%d",&t);
	while(t--) {
		scanf("%d%d%d%lf",&n,&m,&k,&p);
		printf("Case %d: %f\n",++cas,n*p*k);
	}
	return 0;
}

总结：理解题意，注意输出是%f。

题目链接