HDU 4586 Play the dice(概率题，推公式)

Problem Description:

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 ai on it. Now here is a game that you can roll this dice once, if the i-th side is up, you will get
ai 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.

Input:

Input consists of multiple cases. Each case includes two lines.

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.

Output:

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.

Sample Input:

6 1 2 3 4 5 6

0

4 0 0 0 0

1 3

Sample Output:

3.50

0.00

解题思路：

当n == m时，输出inf, 不然结果就是 sum / (n - m);

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define LL long long
using namespace std;
const int MAXN = 1000 + 10;
int main()
{
    int sum, n, m;
    while(scanf("%d", &n)!=EOF)
    {
        sum = 0;
        int x;
        for(int i=0;i<n;i++)
        {
            scanf("%d", &x);
            sum += x;
        }
        scanf("%d", &m);
        for(int i=0;i<m;i++)
            scanf("%d", &x);
        if(sum == 0)
        {
            printf("0.00\n");
            continue;
        }
        if(n == m)
        {
            printf("inf\n");
            continue;
        }
        printf("%.2lf\n", (double)sum / (n - m));
    }
    return 0;
}