LightOJ 1027 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 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

题意：有n个门, 正数表示经过时间 t 能走出迷宫, 负数表示经过时间 t 返回到原来位置（不仅出不去，还原地踏步t时间）；
思路：设门的总数是N, 有N1个门可以走出迷宫, N2个门会返回原来的位置. T1为总的总出迷宫的时间, T2为总的返回原位的时间，则 P(走出迷宫)= N1 / N , P(返回)= N2 / N ；
期望实验次数（走出去的期望次数）: E(X)= N / N1 ；
每次实验消耗的时间: T= （T1+T2）/ N ；
所需时间的数学期望: E(t)=(N / N1) * T = ( T1 + T2) / N1 ;（次数 * 单次花费时间）
期望次数的大神推导：

#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

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

int main() {
    int t, p =1;
    scanf("%d", &t);
    while(t--) {
        int n, k;
        int sum = 0, ans = 0;;
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            scanf("%d", &k);
            if(k > 0) ans++;
            sum += abs(k);
        }
        printf("Case %d: ", p++);
        if(ans) printf("%d/%d\n", sum / gcd(sum, ans), ans / gcd(sum, ans)); //注意化简 
        else puts("inf");
    }
    return 0;
}