[LOJ 1027] Dangerous Maze

A - A Dangerous Maze

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Description

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

3

1

1

2

-10 -3

3

3 -6 -9

Sample Output

Case 1: 1/1

Case 2: inf

Case 3: 18/1

貌似是第一道关于期望和概率的题，唉、弱

分析：设出去的时间期望等于(E)，出去分为两种情况：
A. 一次就出去了,则(P1=n1/n),(n1)表示正数的个数,平均时间(T1=SUM(ai)/n1),(ai)为正数；
B. 第一次没出去,则(P2=n2/n),(n2)表示负数的个数,平均时间为回到起点的平均时间+
从起点出去的平均时间,前者(T2=SUM(ai)/n2),(ai)为负数,后者即为(E)；
综上：(E=P1*T1+P2*(T2+E))
解得：(E=(P1*T1+P2*T2)/(1-P2))

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
#define N 110

int main()
{
    int T,iCase=1;
    int n,n1,n2;
    int sum1,sum2;
    scanf("%d",&T);
    while(T--)
    {
        n1=n2=0;
        sum1=sum2=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            if(x>0)
            {
                n1++;
                sum1+=x;
            }
            else
            {
                n2++;
                sum2-=x;
            }
        }
        int k1=sum1+sum2;
        int k2=n-n2;
        int k=__gcd(k1,k2);
        printf("Case %d: ",iCase++);
        if(k2==0)
            printf("inf
");
        else
            printf("%d/%d
",k1/k,k2/k);
    }
    return 0;
}