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.



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

#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;
}