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+;
 const double oula=0.57721566490153286060651209;
 using namespace std;

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

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