LightOJ1027---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 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

Output for Sample Input

3

1

1

2

-10 -3

3

3 -6 -9

Case 1: 1/1

Case 2: inf

Case 3: 18/1

Problem Setter: Jane Alam Jan

首先设走出去的期望为$E$, 走一次就出去的概率为$p$, 走一次出去的平均时间为$T1$,走一次回到原点的概率为$q$,平均时间为$T2$
则有公式:
$E = p \cdot T1 + q \cdot(T2 + E)$
如果$p 不为0$ 就有解
$(1 - q) \cdot E = p \cdot T1 + q \cdot T2$
$p \cdot E = p \cdot T1 + (1 - p) \cdot T2$
设可以出去的门的时间和为$sumT1$, 回到原点的门的时间和为$sumT2$, 可以出去的门有m扇
所以上式变为:
$m\over n$$\cdot E =$ $m \over n$ $\cdot$ $sumT1 \over m$ $+$ $n - m \over n$ $\cdot$ $sumT2 \over n - m$

所以: $m\over n$$\cdot E =$ $sumT1 + sumT2 \over n$
所以: $E =$ $sumT1 + sumT2 \over m$

/*************************************************************************
    > File Name: A.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月17日 星期日 15时23分53秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

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

int main() {
    int t;
    int icase = 1;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        int all = 0;
        int cnt = 0;
        int u;
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &u);
            all += abs(u);
            if (u > 0) {
                ++cnt;
            }
        }
        if (!cnt) {
            printf("Case %d: inf\n", icase++);
        }
        else {
            u = gcd(all, cnt);
            printf("Case %d: %d/%d\n", icase++, all/ u, cnt / u);
        }
    }
    return 0;
}