LightOj-1027 A Dangerous Maze(期望)

You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors.

If you choose the i^th door, it can either take you back to the same position where you begun in x_i minutes, or can take you out of the maze after x_i minutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience.

Now you want to find the expected time to get out of the maze.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a blank line and an integer n (1 ≤ n ≤ 100) denoting the number of doors. The next line contains nspace separated integers. If the i^th integer (x_i) is positive, you can assume that the i^th door will take you out of maze after x_i minutes. If it's negative, then the i^th door will take you back to the beginning position after abs(x_i) minutes. You can safely assume that 1 ≤ abs(x_i) ≤ 10000.

Output

For each case, print the case number and the expected time to get out of the maze. If it's impossible to get out of the maze, print 'inf'. Print the result in p/q format. Where p is the numerator of the result and q is the denominator of the result and they are relatively prime. See the samples for details.

Sample Input

3

 

1

1

 

2

-10 -3

 

3

3 -6 -9

Sample Output

Case 1: 1/1

Case 2: inf

Case 3: 18/1

题解：给你n扇门，每扇门都有一个数，k代表从这扇门走k分钟出去，-k代表从这扇门走-k分钟回到原点，让你求出去的期望；

设期望为d 则有(以样例3位例子)：d=(3+6+d+9+d)/3;

参考代码为：

#include<bits/stdc++.h>
using namespace std;
int T,n,a[110],sum,cnt;

int gcd(int a,int b){ return b==0? a:gcd(b,a%b); }

int main()
{
    scanf("%d",&T);
    for(int k=1;k<=T;k++)
    {
        sum=0;cnt=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",a+i),sum+=abs(a[i]);
            if(a[i]<0) cnt++;
        }
        if(cnt==n) printf("Case %d: inf\n",k);
        else
        {
            cnt=n-cnt;
            if(cnt>sum) swap(cnt,sum);
            int num=gcd(sum,cnt);
            //int num= __gcd(sum,cnt);
            printf("Case %d: %d/%d\n",k,sum/num,cnt/num);
        }
    }
    
    return 0;
}