1737 Play the Dice （期望 ）

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;
}