Problem:Given the N integers, you have to find the maximum GCD (greatest common divisor) of every possible pair of these integers.
Input :The first line of input is an integer N (1 < N < 100) that determines the number of test cases. The following N lines are the N test cases. Each test case contains M (1 < M < 100) positive integers that you have to find the maximum of GCD.
Output :For each test case show the maximum GCD of every possible pair.
Sample Input 3 10 20 30 40 7 5 12 125 15 25 Sample Output 20 1 25
题解：读入的时候处理一下，可以直接读入一个字符串，然后把数再按十进制还原存到数组中，或者直接用ungetc来退回一下。
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int a[150];
int main()
{
int n, maxx = -1;
char op;
while(~scanf("%d",&n))
{
while(n --)
{
maxx = -1;
int i = 0;
while(1)
{
scanf("%d",&a[i ]);
while((op=getchar())==' '); // 如果是空格的话用ungetc退格
ungetc(op, stdin);
if(op == 'n') break;
}
for(int j = 0; j < i; j )
{
for(int k = j 1; k < i; k )
{
if(__gcd(a[j],a[k]) > maxx) maxx = __gcd(a[j],a[k]);
}
}
printf("%dn",maxx);
}
}
return 0;
}
