1737 - Play the Dice （期望）

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最新推荐文章于 2025-02-09 10:00:00 发布

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本文介绍了一个基于概率论的游戏算法，通过计算不同情况下玩家在特殊规则的骰子游戏中可能获得的平均收益，探讨了数学期望在游戏设计中的应用。

1737 - Play the Dice

http://icpc.njust.edu.cn/Local/1737

时间限制:	2000 MS
内存限制:	65535 MB

问题描述

There is a dice with N sides, which are numbered from 1,2,...,n and have the equal possibility to show up when one rolls a dice. Each side has an integer A _i on it. Now here is a game that you can roll this dice once, if the i-th side is up, you will get A _i yuan. What's more, some sids of this dice are colored with a special different color. If you turn this side up, you will get once more chance to roll the dice. When you roll the dice for the second time, you still have the opportunity to win money and rolling chance. Now you need to calculate the expectations of money that we get after playing the game once.

输入说明

Input consists of multiple cases. Each case includes two lines. End with EOF.
The first line is an integer N (2<=N<=200), following with N integers A _i(0<=A _i<200)
The second line is an integer M (0<=M<=N), following with m integers B _i(1<=B _i<=n), which are the numbers of the special sides to get another more chance.

输出说明

Just a real number which is the expectations of the money one can get, rounded to exact two digits. If you can get unlimited money, print "inf" a line without double quotes.

输入样例

6 1 2 3 4 5 6
0
4 0 0 0 0
1 3

输出样例

3.50
0.00

来源

2013 ACM-ICPC China Nanjing Invitational Programming Contest

ans=M/N *ans+1/N*(A[1]+A[2]+A[3]+....+A[N]);

(N-M)ans= A[1]+A[2]+...+A[N]=sum;

如果sum==0,答案为0.00

如果sum!=0,N-M==0 答案为inf

否则答案就是sum/(N-M);

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=220;

int a[N],b[N];
int n,m;

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n)){
        int sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        //printf("sum=%d\n",sum);
        scanf("%d",&m);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        if(sum==0){
            printf("0.00\n");
            continue;
        }
        if(n==m){
            printf("inf\n");
            continue;
        }
        printf("%.2lf\n",1.0*sum/(n-m));
    }
    return 0;
}