Codeforces Round #720 (Div. 2) A. Nastia and Nearly Good Num

Posted 2021-06-21 小哈里

problem

A. Nastia and Nearly Good Numbers
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Nastia has 2 positive integers A and B. She defines that:

The integer is good if it is divisible by A⋅B;
Otherwise, the integer is nearly good, if it is divisible by A.
For example, if A=6 and B=4, the integers 24 and 72 are good, the integers 6, 660 and 12 are nearly good, the integers 16, 7 are neither good nor nearly good.

Find 3 different positive integers x, y, and z such that exactly one of them is good and the other 2 are nearly good, and x+y=z.

Input
The first line contains a single integer t (1≤t≤10000) — the number of test cases.

The first line of each test case contains two integers A and B (1≤A≤106, 1≤B≤106) — numbers that Nastia has.

Output
For each test case print:

“YES” and 3 different positive integers x, y, and z (1≤x,y,z≤1018) such that exactly one of them is good and the other 2 are nearly good, and x+y=z.
“NO” if no answer exists.
You can print each character of “YES” or “NO” in any case.
If there are multiple answers, print any.

Example
inputCopy
3
5 3
13 2
7 11
outputCopy
YES
10 50 60
YES
169 39 208
YES
28 154 182
Note
In the first test case: 60 — good number; 10 and 50 — nearly good numbers.

In the second test case: 208 — good number; 169 and 39 — nearly good numbers.

In the third test case: 154 — good number; 28 and 182 — nearly good numbers.

solution

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
LL gcd(LL a, LL b){return !b?a:gcd(b,a%b);}
LL lcm(LL a, LL b){ return a/gcd(a,b)*b; }
int main(){
	int T;  cin>>T;
	while(T--){
		LL a, b;  cin>>a>>b;
		if(b==1){cout<<"NO\\n"; continue;}
		cout<<"YES\\n";
		cout<<a*(b-1)<<" "<<a*(b+1)<<" "<<a*(b)*2<<"\\n";
	}
	
	return 0;
}