Codeforces Round #656 (Div. 3)题解总结


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
Codeforces Round 894 (Div. 3) 是一个Codeforces举办的比赛,是第894轮的Div. 3级别比赛。它包含了一系列题目,其中包括题目E. Kolya and Movie Theatre。 根据题目描述,E. Kolya and Movie Theatre问题要求我们给定两个字符串,通过三种操作来让字符串a等于字符串b。这三种操作分别为:交换a中相同位置的字符、交换a中对称位置的字符、交换b中对称位置的字符。我们需要先进行一次预处理,替换a中的字符,然后进行上述三种操作,最终得到a等于b的结果。我们需要计算预处理操作的次数。 根据引用的讨论,当且仅当b[i]==b[n-i-1]时,如果a[i]!=a[n-i-1],需要进行一次操作;否则不需要操作。所以我们可以遍历字符串b的前半部分,判断对应位置的字符是否与后半部分对称,并统计需要进行操作的次数。 以上就是Codeforces Round 894 (Div. 3)的简要说明和题目E. Kolya and Movie Theatre的要求。<span class="em">1</span><span class="em">2</span><span class="em">3</span> #### 引用[.reference_title] - *1* *2* [Codeforces Round #498 (Div. 3) (A+B+C+D+E+F)](https://blog.csdn.net/qq_46030630/article/details/108804114)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"] - *3* [Codeforces Round 894 (Div. 3)A~E题解](https://blog.csdn.net/gyeolhada/article/details/132491891)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"] [ .reference_list ]
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