zoj 3844 Easy Task(水题)

题目链接

Easy Task

Time Limit: 2 Seconds
Memory Limit: 65536 KB

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 b with 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

题意：n个数字，每次操作找出最大的数和最小的数，并把他们替换成（最大数和最小数之差）。一直操作，直至所有数字都相同。输出最后的数字。如果无解输出NOOOOO！。

题解：显然是没有无解的情况的，由于n<=10。所以直接暴力就好了。

代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<stdlib.h>
#include<vector>
#define inff 0x3fffffff
#define nn 1100
#define mod 1000000007
typedef long long LL;
const LL inf64=inff*(LL)inff;
using namespace std;
int n;
int a[15];
int main()
{
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int mx,mn,id1,id2;
while(1)
{
mx=-inff,mn=inff;
for(i=1;i<=n;i++)
{
if(a[i]>mx)
{
mx=a[i];
id1=i;
}
if(a[i]<mn)
{
mn=a[i];
id2=i;
}
}
if(mn==mx)
break;
a[id1]=mx-mn;
a[id2]=mx-mn;
}
printf("%d\n",mx);
}
return 0;
}