题目描述
Nastia has 22 positive integers AA and BB . She defines that:
The integer is good if it is divisible by A \cdot BA?B ;
Otherwise, the integer is nearly good, if it is divisible by AA .
For example, if A = 6A=6 and B = 4B=4 , the integers 2424 and 7272 are good, the integers 66 , 660660 and 1212 are nearly good, the integers 1616 , 77 are neither good nor nearly good.
Find 33 different positive integers xx , yy , and zz such that exactly one of them is good and the other 22 are nearly good, and x + y = zx+y=z .
输入格式
The first line contains a single integer tt ( 1 \le t \le 10,0001≤t≤10000 ) — the number of test cases.
The first line of each test case contains two integers AA and BB ( 1 \le A \le 10^61≤A≤10
6
, 1 \le B \le 10^61≤B≤10
6
) — numbers that Nastia has.
输出格式
For each test case print:
“YES” and 33 different positive integers xx , yy , and zz ( 1 \le x, y, z \le 10^{18}1≤x,y,z≤10
18
) such that exactly one of them is good and the other 22 are nearly good, and x + y = zx+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.
题意翻译
题目描述
Nastia有两个正整数 A,BA,B
如果一个正整数 CC 可以被 A*BA?B 整除，则称这个正整数CC为“好数”。
如果一个正整数 CC 不能被 A*BA?B 整除却能被AA整除，则称这个正整数CC为“接近好数”。
请你找出三个正整数 x,y,zx,y,z, 使他们中恰好有一个数是“好数”且另外两个数是“接近好数”，且满足x+y=zx+y=z。
输入格式
第一行一个正整数tt, 表示测试数据的组数。
接下来tt行，每行两个正整数A,BA,B, 如题意所述。
输出格式
对于每一组数据，输出"YES"或"NO",表示是否能找到符合要求的x,y,zx,y,z。如果可以找到答案，在下一行输出任意一组符合要求的x,y,zx,y,z，以空格隔开。
输入输出样例
输入
3
5 3
13 2
7 11
输出
YES
10 50 60
YES
169 39 208
YES
28 154 182
说明/提示
In the first test case: 6060 — good number; 1010 and 5050 — nearly good numbers.
In the second test case: 208208 — good number; 169169 and 3939 — nearly good numbers.
In the third test case: 154154 — good number; 2828 and 182182 — nearly good numbers.
//只要保证总长度是偶数，且首尾合法，就一定可以，因为给的字符串里只会出现一个（和），其余一定都是问号
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;cin>>n;while(n--){
string s;cin>>s;if(s.length()%2==0&&s[0]!=')'&&s[s.length()-1]!='('){
cout<<"YES"<<endl;}else{
cout<<"NO"<<endl;}}return 0;
}