lightoj-1027 - A Dangerous Maze(数学期望)

1027 - A Dangerous Maze
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Time Limit: 2 second(s) Memory Limit: 32 MB
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

 

题目大意: 给你n个门，每次进每个门的概率都是一样的，正数代表你x分钟后可以离开这里，负数代表你x分钟后回到这里。求预期离开的时间。

思路: 大概就是求离开这里所需要的期望值。

有题目可知，离开这里的方法是无穷多种，所以一般的期望公式是没办法列出来的

我们先设离开这里的期望值是E;

很明显E=(t1+t2+……+tn + G1+G2+……+Gm)/(n+m);ps:t代表直接选到正数离开，G代表包含负数 离开迷宫的情况

负数想要构成离开的条件就必须要和一个可以离开的情况构成搭配，则Gm = abs(NEGATIVEm) + E;

则明显E = (t1+t2+……+tn + (abs(T1)+E)+(abs(T2)+E)+……+(abs(Tm)+E))/n → E = (t1+t2+……+tn+abs(T1)+abs(T2)+……+abs(Tm))/(n-m);

特殊情况 n-m =0 时，答案为inf

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

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

int main(){
    
    int T,sum,nn,n,m,temp;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        sum = m = 0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&temp);
            if(temp<0) m++,temp = abs(temp);
            sum += temp;
        }
        nn = n-m;
        if(nn==0) printf("Case %d: inf\n",t);
        else{            
            int gg = gcd(nn,sum);
            printf("Case %d: %d/%d\n",t,sum/gg,nn/gg);            
        }        
    } 
    
    return 0;
}