LightOJ-1027-A Dangerous Maze(概率）

链接：

https://vjudge.net/problem/LightOJ-1027#author=634579757

题意：

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.

思路：

考虑当x大于0则出去的期望时间为1/nt,如果小于0出去的期望时间为1/n(t+d).其中d为整体出去的期望时间。

推出d = 1/nsumt+ne/nd.其中sumt为总时间及abs(t)的和。ne为负值的们。

代码：

#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
#include <set>
#include <iterator>
#include <cstring>
#include <assert.h>
using namespace std;

typedef long long LL;

int n;

int main()
{
    int t, cnt = 0;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d", &n);
        int sum = 0, ne = 0, v;
        bool flag = false;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d", &v);
            if (v > 0)
                flag = true;
            if (v < 0)
                ne++;
            sum += abs(v);
        }
        if (!flag)
            printf("Case %d: inf\n", ++cnt);
        else
            printf("Case %d: %d/%d\n", ++cnt, sum/__gcd(sum, n-ne), (n-ne)/__gcd(sum, n-ne));
    }

    return 0;
}