Codeforces Round #656 (Div. 3)题解总结

最新推荐文章于 2024-08-31 12:42:46 发布

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最新推荐文章于 2024-08-31 12:42:46 发布 · 1.1k 阅读

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文章目录

A. Three Pairwise Maximums
B. Restore the Permutation by Merger
C. Make It Good
D. a-Good String

来源： http://codeforces.com/contest/1385

A. Three Pairwise Maximums

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

题意：
给你三个正整数x、y和z，请你找到正整数a，b和c，使得x=max(a，b)，y=max(a，c)且z=max(b，c)，或者确定不可能找到这样的a，b和c(你可以以任意顺序输出a,b,c）

思路：
因为x=max(a，b)，y=max(a，c)x且z=max(b，c)，我们可以得出x,y,z中至少有两个数是相等的，且一定大于等于另一个数
如果x=y则说明a为最大值，此时需要满足a>=z，如果不满足该条件，则无解，因为z=max(b，c)，我们不能确定b,c谁比较大，所以我们就假设两个数一样的即可。
x=z时和y=z时同理

代码：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<cstdio>
using namespace std;
typedef long long ll;
int inf = 0x3f3f3f3f;
const ll N = 2e6+5;
int main()
{
   
   
    int t;
    scanf("%d", &t);
    while (t--)
    {
   
   
		int a = 0, b = 0, c = 0, x, y, z;
		cin >> x >> y >> z;
		if (x == y) a = x;
		else if (y == z) c = y;
		else if (x == z) b = x;
		if ((a && a >= z) || (b && b >= y) || (c && c >= x)) 
		{
   
   
			cout << "YES" << endl;
			if (a) cout << a << " " << z << " " << z << endl;
			else if (b) cout << b << " " << y << " " << y << endl;
			else if (c)