Maximum GCD(fgets读入)

Maximum GCD

https://vjudge.net/contest/288520#problem/V

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

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 10000005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<long long,int>pli;
typedef pair<int,char> pic;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long mod=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

int n;
char str[10005];
vector<int>ve;

int main(){
    scanf("%d%*c",&n);
    for(int i=1;i<=n;i++){
        fgets(str,sizeof(str),stdin);
        ve.clear();
        int co=0;
        int len=strlen(str);
        for(int j=0;j<len;j++){
            if(str[j]>='0'&&str[j]<='9'){
                co=co*10+str[j]-'0';
            }
            else{
                if(co)
                    ve.pb(co);
                co=0;
            }
        }
        if(co){
            ve.pb(co);
        }
        int ans=1;
        for(int i=0;i<ve.size();i++){
            for(int j=i+1;j<ve.size();j++){
                ans=max(ans,__gcd(ve[i],ve[j]));
            }
        }
        printf("%d
",ans);
    }
}