Maximum GCD

Given the
Nintegers, you have to nd the maximum GCD (greatest common divisor) of every possiblepair of these integers.
Input
The rst line of input is an integerN(1

这里写代码片#include<iostream>#include<algorithm>#include<cstring>#include<string>using namespace std;int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int main(){    int t;    scanf("%d",&t);    getchar();//避免跳行被读入    while(t--)    {        int a[110],sum=0,len=0,maxn=0;        char s;        while(1)        {            scanf("%c",&s);            if(s=='\n')            break;            if(s>='0'&&s<='9')            sum=sum*10+s-'0';            else            {                if(sum==0)//空格连续                continue;                if(sum!=0)                a[len++]=sum;                sum=0;            }        }        if(sum!=0)        a[len++]=sum;        for(int i=0;i<len;i++)        {            for(int j=i+1;j<len;j++)            {                maxn=max(maxn,gcd(a[i],a[j]));            }        }        printf("%d\n",maxn);    }    return 0; }