（期望）A Dangerous Maze（Light OJ 1027）

http://www.lightoj.com/volume_showproblem.php?problem=1027

 

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	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

题目大意：

 

在迷宫里， 你现在在起点， 你面前有n扇门，每个门上有一个数字表示这个门x分钟后会打开，负数代表这个门-x分钟后会回到起点，选择门的概率是一样的并且每次互不影响，问出去的时间期望是多少

（具体推导云笔记上写的有）

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef unsigned long long LL;
#define met(a,b) (memset(a,b,sizeof(a)))
const int INF = 1e9+7;
const int maxn = 105;
const int MOD = 9973;

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

int main()
{
    int T, iCase=1;
    scanf("%d", &T);
    while(T--)
    {
        int n, time, i, sum=0, k=0;

        scanf("%d", &n);

        for(i=1; i<=n; i++)
        {
            scanf("%d", &time);
            if(time>0) k++;
            sum += abs(time);
        }

        int w = gcd(sum, k);

        if(!k)
            printf("Case %d: inf\n", iCase++);
        else
            printf("Case %d: %d/%d\n", iCase++, sum/w, k/w);

    }
    return 0;
}
/**

*/

 

参考  http://www.cnblogs.com/WABoss/p/5296560.html

 

转载于:https://www.cnblogs.com/YY56/p/5533744.html