LightOJ 1027 A Dangerous Maze (期望)

题目链接：LightOJ - 1027

Description

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 n space 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

Solution

题意

你在一个迷宫里，面前有 (n) 扇门，如果第 (i) 扇门的 (X_i) 值为正值，就可以花费 (X_i) 的时间的走出迷宫，否则花费 (abs(X_i)) 的时间又回到原点，且不记得之前走过哪些门。每次等概率选择一扇门，问走出迷宫的时间的期望。不能走出去输出 (inf)。

思路

设有 (n_1) 扇门能走出迷宫，(n_2) 不能走出迷宫，则 (n_1 + n_2 = n)。

设 (n_1) 扇门的 (X_i) 值的平均值为 (t_1)，(n_2) 扇门的 (X_i) 值的平均值为 (t_2).

设走出去的期望为 (E)。则 (E = frac{n_1}{n} cdot t_1 + frac{n_2}{n} cdot (t_2 + E))。

化简后为 (E = frac{1}{n_1}sum_{i=1}^n abs(X_i))。

Code

#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;

int x[maxn];

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    int kase = 0;
    while(T--) {
        int n;
        cin >> n;
        int sum = 0;
        int cnt = 0;
        for(int i = 1; i <= n; ++i) {
            cin >> x[i];
            if(x[i] > 0) {
                ++cnt;
                sum += x[i];
            } else {
                sum += -x[i];
            }
        }
        int d = gcd(sum, cnt);
        if(cnt == 0) printf("Case %d: inf
", ++kase);
        else printf("Case %d: %d/%d
", ++kase, sum / d, cnt / d);
    }
    return 0;
}