Easy Task

You are given n integers. Your task is very easy. You should find the maximum integer a and the minimum integer b among these n integers. And then you should replace both a and bwith a-b. Your task will not be finished unless all the integers are equal.

Now the problem come, you want to know whether you can finish you task. And if you can finish the task, you want to know the final result.

Input

The first line of the input contain an integer T(T≤ 20) indicates the number of test cases.

Then T cases come. Each case consists of two lines. The first line is an integer n(2≤ n≤ 10) as the problem described. The second line contains n integers, all of them are no less than -100000 and no more than 100000.

Output

For each case you should print one line. If you can finish your task, you should print one of the n integers. Otherwise, you should print "Nooooooo!"(without quotes).

Sample Input

2
3
1 2 3
2
5 5

Sample Output

2
5

//最大值与最小值相减，使最后所有数都相等。
//循坏排序得出两端的最大值与最小值，相减。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
  int t,n,sum;
  int a[12],i;
  while(scanf("%d",&t)!=EOF)
  {
    while(t--)
    {
    scanf("%d",&n);
    for(i=0;i<n;i++)
      scanf("%d",a+i);
      
    while(1)
    {
      sort(a,a+n);
      if(a[n-1]==a[0]) break;//最大值与最小值相等，则所有数都相等。
      sum=a[n-1]-a[0];
      a[n-1]=a[0]=sum;
    }                 
     printf("%d\n",a[n-1]);      
  }   
}
return 0;
}

Easy Task

原文：http://blog.csdn.net/u012346225/article/details/43636603