LightOJ 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 i^th door, it can either take you back to the same position where you begun inx_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 containsn 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 afterx_i minutes. If it's negative, then the i^th door will take you back to the beginning position afterabs(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

3

 

1

1

 

2

-10 -3

 

3

3 -6 -9

Sample Output

Case 1: 1/1

Case 2: inf

Case 3: 18/1

题意是这样的：

你身处迷宫中，面前有n个门，每次都可以任选一个门，如果门的时间是正数，则表示t时间后会走出迷宫，反之表示t时间后返回原地，返回原地后不会记得之前的选择，还要重新选择，求走出迷宫的时间期望，如果走不出去输出“inf”；

有两种情况：

（1）选中的门的t为正，则期望为t/n；

（2）选中的门t为负，则回到原地期望为t/n，出去的期望为（t+E）/n；

设sum1，sum2分别为t为正的门的时间总和和t为负的门的时间总和；door1，door2分别为t为正的门的个数和t为负的门的个数；

则E=sum1/N+（sum2+door2*E）/N;化简得E=（sum1+sum2）/door1；

下面是代码：

#include <iostream>using namespace std;int gcd(int a, int b){    if(a<b){        a=a+b;        b=a-b;        a=a-b;    }    return a%b==0?b:gcd(b, a%b);}int main(){    int T;    cin >> T;    int cnt=0;    while(T--){        cnt++;        int sum1, sum2, door1, door2;        sum1=sum2=door1=door2=0;        int n;        cin >> n;        for(int i=1; i<=n; i++){            int x;            cin >> x;            if(x>0){                sum1+=x;                door1++;            }            else{                sum2+=(-x);                door2++;            }        }        cout << "Case " << cnt << ": ";        if(door1==0) cout << "inf\n";        else{            int x=(sum1+sum2)/gcd(sum1+sum2, door1);            int y=door1/gcd(sum1+sum2, door1);            cout << x << '/' << y << endl;        }    }    return 0;}