A Dangerous Maze LightOJ

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 ith door, it can either take you back to the same position where you begun in xi minutes, or can take you out of the maze after xi 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 n space separated integers. If the ith integer (xi) is positive, you can assume that the ith door will take you out of maze after xi minutes. If it’s negative, then the ith door will take you back to the beginning position after abs(xi) minutes. You can safely assume that 1 ≤ abs(xi) ≤ 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

/*如果大于0  期望为1 / N * T如果小于0  期望为1 / N * T。回到原来位置，所以期望为1 / N * (T + E)求和    E=1 / N * (大于0的和)  + 1 / N * (小于0的绝对值和 + 小于0的个数 * E)整理    E = (全部数绝对值的和) / (N-小于0 的个数)*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int gcd(int a, int b){    return b ? gcd(b, a%b) : a;}int main(){    int t;    int n;    int q=1;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        int a;        int sum=0;        int x=0;        for (int i=0;i<n;i++){            scanf("%d",&a);            if (a<0){                sum+=-a;            }            else{                x++;                sum+=a;            }        }        int xx=gcd(sum,x);        printf("Case %d: ",q++);        if (x==0){            printf("inf\n");        }        else        printf("%d/%d\n",sum/xx,x/xx);    }    return 0;}