Algorithm
You are given a positive integer x. Find any such 2 positive integers a and b such that GCD(a,b)+LCM(a,b)=x.
As a reminder, GCD(a,b) is the greatest integer that divides both a and b. Similarly, LCM(a,b) is the smallest integer such that both a and b divide it.
It's guaranteed that the solution always exists. If there are several such pairs (a,b), you can output any of them.
The first line contains a single integer t (1≤t≤100) — the number of testcases.
Each testcase consists of one line containing a single integer, x (2≤x≤109).
For each testcase, output a pair of positive integers a and b (1≤a,b≤109) such that GCD(a,b)+LCM(a,b)=x. It's guaranteed that the solution always exists. If there are several such pairs (a,b), you can output any of them.
2 2 14
1 1 6 4
In the first testcase of the sample, GCD(1,1)+LCM(1,1)=1+1=2.
In the second testcase of the sample, GCD(6,4)+LCM(6,4)=2+12=14.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--){
int n;
cin>>n;
cout<<1<<" "<<(n-1)<<endl;
}
}Input
2
14
Output
6 4
Demonstration
Codeforcess solution 1325-A, A. EhAb AnD gCd, ,C++, Java, Js and Python,1325-A,Codeforcess solution