1027 - A Dangerous Maze - dblank

1027 - A Dangerous Maze

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.
<h1>Input</h1>
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 ithdoor 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.
<h1>Output</h1>

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.

<h1>Sample Input</h1> <h1>Output for Sample Input</h1>
3   1 1   2 -10 -3   3 3 -6 -9 Case 1: 1/1 Case 2: inf Case 3: 18/1

题意:

有n扇门,每次你可以选择其中一扇。xi为负值的门带你abs(xi)后又回到原点。xi为正值的门则带你在xi后离开迷宫。每次你都没有经验没有记忆。选择每扇门的概率相等。求走出迷宫的时间期望值。

思路:只要选到了正值门则一定可以出去,选负值门则会回到原点。因为出去的话,最后一次则一定是正值门,我们看到选负值门的次数可能是0到正无穷大次,每次给期望时间增加的是所有负值门的平均值的绝对值,这个可以是一个以选到负值门的概率的等比数列的前n项和,化简一下就好了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f, N = 100 + 10, Mod = 1000000007;
typedef long long ll;
int gcd(int a, int b)
{
    return b == 0 ?a: gcd(b, a%b);
}
int p[N];
int main()
{
   // freopen("d:\input.txt", "r", stdin);
  //  freopen("d:\output.txt", "w", stdout);
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas<=t; cas++)
    {
        int n, cnt = 0, sum1 = 0, sum2 = 0;
        scanf("%d", &n);
        for(int i = 0; i<n; i++)
        {
            scanf("%d", &p[i]);
            if(p[i]<0)
                cnt++, sum2 += - p[i];
            else sum1 += p[i];

        }
        printf("Case %d: ", cas);
        if(cnt == n)
        {
            printf("infn");
            continue;
        }
        int ans1 = sum1 + sum2, ans2 = n - cnt;
        printf("%d/%dn", ans1 / gcd(ans1, ans2), ans2/ gcd(ans1, ans2));
    }
    return 0;
}

 

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