A - A Dangerous Maze(期望)

最新推荐文章于 2019-05-26 21:52:12 发布

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本文介绍了一种算法，用于计算从迷宫中逃脱所需的期望时间。输入包括多个测试用例，每个用例包含一定数量的门，每扇门对应着回到起点或直接逃出的时间。算法考虑了所有可能性来计算最终的期望时间，并以分数形式输出。

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

-10 -3

3 -6 -9
Sample Output

Case 1: 1/1

Case 2: inf

Case 3: 18/1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int gcd(int a, int b)
{
   int temp;
   if(a<b)
   {
     temp = a, a = b, b = temp;
   }
   if(b==0)
   return a;
   return gcd(b, a%b);
}
int main()
{
  int T, n, k;
  scanf("%d", &T);
  for(int t=1;t<=T;t++)
  {
    scanf("%d", &n);
    int sum = 0, ans = 0;
    for(int i=0;i<n;i++)
    {
      scanf("%d", &k);
      if(k<0)
      sum -= k;
      else
      {
        sum += k;
        ans++;
      }
    }
    if(ans==0)
    {
       printf("Case %d: inf\n", t);
    }
    else
    {
      printf("Case %d: %d/%d\n", t, sum/gcd(sum, ans), ans/gcd(sum, ans));
    }
  }
  return 0;
}