Omkar, having a laudably curious mind, immediately thought of a problem involving theLCMLCMoperation: given an integernn, find positive integersaaandbbsuch thata+b=na+b=nandLCM(a,b)LCM(a,b)is the minimum value possible.

Can you help Omkar solve his ludicrously challenging math problem?

Each test contains multiple test cases. The first line contains the number of test casestt(1≤t≤101≤t≤10). Description of the test cases follows.

Each test case consists of a single integernn(2≤n≤1092≤n≤109).

For each test case, output two positive integersaaandbb, such thata+b=na+b=nandLCM(a,b)LCM(a,b)is the minimum possible.

3
4
6
9

2 2
3 3
3 6

For the first test case, the numbers we can choose are1,31,3or2,22,2.LCM(1,3)=3LCM(1,3)=3andLCM(2,2)=2LCM(2,2)=2, so we output

For the second test case, the numbers we can choose are1,51,5,2,42,4, or3,33,3.LCM(1,5)=5LCM(1,5)=5,LCM(2,4)=4LCM(2,4)=4, andLCM(3,3)=3LCM(3,3)=3, so we output3333.

For the third test case,LCM(3,6)=6LCM(3,6)=6. It can be shown that there are no other pairs of numbers which sum to99that have a lowerLCMLCM.

題目大意：

給你一個n，找到兩個a,b,使a+b=n,並且lcm(a,b)儘可能小。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int t,x;
 4 bool prim(int x) {
 5     if(x==2)
 6         return true;
 7     if(x<=1)
 8         return false;
 9     for(int i=2; i*i<=x; i++)
10         if(x%i==0)
11             return false;
12     return true;
13 }
14 int gcd(int x) {
15     int ans=-1;
16     for(int i=2; i<x; i++)
17         if(x%i==0)
18         {
19             ans=i;
20             break;
21         }
22     ans=x/ans;
23     return ans;
24 }
25 int main() {
26     scanf("%d",&t);
27     while(t--) {
28         scanf("%d",&x);
29         if(x%2==0)
30             printf("%d %d\n",x/2,x/2);
31         else {
32             if(prim(x))
33                 cout<<1<<" "<<x-1<<endl;
34             else
35                 printf("%d %d\n",gcd(x),x-gcd(x));
36         }
37     }
38     return 0;
39 }