Omkar and Last Class of Math CodeForces - 1372B rat1300

In Omkar’s last class of math, he learned about the least common multiple, or LCM. LCM(a,b) is the smallest positive integer x which is divisible by both a and b.

Omkar, having a laudably curious mind, immediately thought of a problem involving the LCM operation: given an integer n, find positive integers a and b such that a+b=n and LCM(a,b) is the minimum value possible.

Can you help Omkar solve his ludicrously challenging math problem?

Input
Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤10). Description of the test cases follows.

Each test case consists of a single integer n (2≤n≤109).

Output
For each test case, output two positive integers a and b, such that a+b=n and LCM(a,b) is the minimum possible.

Example
Input
3
4
6
9
Output
2 2
3 3
3 6
Note
For the first test case, the numbers we can choose are 1,3 or 2,2. LCM(1,3)=3 and LCM(2,2)=2, so we output 2 2.

For the second test case, the numbers we can choose are 1,5, 2,4, or 3,3. LCM(1,5)=5, LCM(2,4)=4, and LCM(3,3)=3, so we output 3 3.

For the third test case, LCM(3,6)=6. It can be shown that there are no other pairs of numbers which sum to 9 that have a lower LCM.
找到最小因子，就可以得到最大的倍数，再用整数n-倍数

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=350;
typedef long long ll;
int main(){
    int t;
    cin>>t;
    while(t--){
        ll n,i;
        int flag=0;
        cin>>n;
        if(n%2==0){
            cout<<n/2<<" "<<n/2<<endl;
        }
        else{
            for(i=2;i*i<=n;i++){
                if(n%i==0){
                    cout<<n/i<<" "<<n-n/i<<endl;
                    flag=1;
                    break;
                }
            }
            if(flag==0)
                cout<<1<<" "<<n-1<<endl;
        }
    }
  system("pause");
  return 0;
}