Codeforces 1385A

A. Three Pairwise Maximums

time limit per test 1 second
memory limit per test 256 megabytes
input standard input
output standard output
You are given three positive (i.e. strictly greater than zero) integers x, y and z.

Your task is to find positive integers a, b and c such that x=max(a,b), y=max(a,c) and z=max(b,c), or determine that it is impossible to find such a, b and c.

You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.

Input

The first line of the input contains one integer t (1≤t≤2⋅10⁴) — the number of test cases. Then t test cases follow.

The only line of the test case contains three integers x, y, and z (1≤x,y,z≤10⁹).

Output

For each test case, print the answer:

“NO” in the only line of the output if a solution doesn’t exist;
or “YES” in the first line and any valid triple of positive integers a, b and c (1≤a,b,c≤10⁹) in the second line. You can print a, b and c in any order.
Example
inputCopy
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
outputCopy
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000

题目大意：

给出 t 组测试样例，对于每组测试样例给出三个数x y z，你需要寻找三个数a, b, c使得x = max(a,b) y = max(a,c) z = max(b,c) 如果能找到则输出YES按任意顺序打印abc，如果找不到则输出NO；

解题思路：

如果要找到三个数abc，那么给出的xyz 从大到小排序后一定满足 x = y >= z才能找到abc，因为x y z都是由max(…)得来的，abc一定有一个最大，也就是说xyz至少有两个是相等的。这样就可以推出a的值，让a = x(y)，剩下的那两个一定要小于等于最大的数，才能保证剩余的z能通过max得到。因为数据范围是1-1e9，所以让剩下的两个数都等于z即可，放止发生1-1 = 0 的情况。
AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <vector>
using namespace std;
const int N = 15;
typedef long long ll;
int main()
{
	int n;
	cin >> n;
	while (n--)
	{
		ll a[5];
		cin >> a[0] >> a[1] >> a[2];
		sort(a, a + 3, greater<ll >());
		if (a[0] != a[1])
		{
			cout << "NO" << endl;
			continue;
		}
		cout << "YES" << endl;
		cout << a[0] << " " << a[2] << " " << a[2] << endl;
	}
	return 0;
}