Three Pairwise Maximums【数学签到题】

题目描述：原题链接

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
input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000

思路：

首先三个数分别取最大保证x,y,z至少有两个相等。其次，当x,y相等时意味着a为最大值，b,c取相同的较小值即可，以此类推到后两种情况。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int N = 100010;

int main()
{
	int T;
	scanf("%d",&T);
	while (T--)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        if(x==y && x>=z) printf("YES\n%d %d %d\n",x,z,z);
        else if(y==z && z>=x) printf("YES\n%d %d %d\n",x,x,z);
        else if(z==x && z>=y) printf("YES\n%d %d %d\n",y,z,y);
        else puts("NO");
	}
	return 0;
}