798A - Mike and palindrome

最新推荐文章于 2022-10-19 20:41:21 发布

AndrewThompson

最新推荐文章于 2022-10-19 20:41:21 发布

阅读量384

点赞数

CC 4.0 BY-SA版权

分类专栏： Codeforces

本文链接：https://blog.youkuaiyun.com/zju2016/article/details/70665953

Codeforces 专栏收录该内容

32 篇文章

订阅专栏

本文介绍了一个简单的字符串处理问题，即如何判断一个字符串是否可以通过仅改变一个字符而变为回文串。通过两端向中间遍历的方法，文章提供了一种有效的算法解决方案，并详细解释了算法的工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

In this problem,we can change the only one character.We define a variable amount to count the number of different characters when we traverse from the both ends to the middle.When the amount>=2,it means we must change more than once,so we just print "NO".When the amount equals one,we just print "YES".But when the amount equals zero,what we should do depends on the size of the string.If the size is odd,we can just change the alphebet in the middle,so in this case,we can print "YES".However, when the size of the string is even,it means we have to change at least two alphebats.So we should print "NO".Well, that is all.

#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<algorithm>
#include<queue>
using namespace std;

int main(){
	string s;
	while (cin >> s){
		int i = 0, j = s.size() - 1;
		int amount = 0;
		while (i < j){
			if (s[i] != s[j]) amount++;
			if (amount >= 2){
				cout << "NO" << endl;
				break;
			}
			i++;
			j--;
		}
		if (amount == 1) cout << "YES" << endl;
		else if (amount == 0){
			if (s.size() % 2)
				cout << "YES" << endl;
			else cout << "NO" << endl;
		}
	}
	system("pause");
	return 0;
}