Solution to a problem CF1325A [EhAb AnD gCd]

\(Solution:\)

This question is the output of a set of questions \ ((a, b) \ ) such that \ (\ GCD (A, B) + \} OperatorName LCM {(A, B) = X \) , we know that \ (\ GCD (A,. 1). 1 = \) , \ (\ OperatorName} {LCM (A,. 1) A = \) and to know if there are several possible outputs to one, so we directly outputted $ 1, x -1 $ on the line.

\(code:\)

#include<cstdio>//这里为了节省空间，所以不打万能头
using namespace std;//标准数据库
inline int read()//快速读入
{
    int x=0;bool f=1;char c=getchar();
    while(c<'0' || c>'9'){if(c=='-') f=0;c=getchar();}
    while(c>='0' && c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
    return f?x:-x;
}
int T,n;
int main()
{
    T=read();//读入数据组数
    while(T--)
    {
        x=read();//读入 x
        printf("%d %d\n",1,x-1);//按以上讲述方法，直接输出1,x-1 即可
    }
    return 0;
}

If this article has given you to help, then you point a praise and then go, Thanks ♪ (· ω ·) Techno