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[CF#712Div2]A题Déjà Vu

作者:互联网

Pro

CF#712Div2.A

Sol

这个题不是很难 思路啥的也很好想 就说几个需要注意的地方吧

字符串切割分开输出的时候一定要注意substr的用法

分析什么时候可以输出NO是解题关键

其余都是一点调试就能找出来的小细节

官方题解

在这里插入图片描述

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define INF 2147483647
#define eps 1e-7
#define L 100005
#define Fo(i,a,b) for(LL i=(a); i<=(b); i++)
#define Ro(i,b,a) for(LL i=(b); i>=(a); i--)
#define Ms(a,b) memset((a),(b),sizeof(a))
#define _ceil(_,__) (_+(__-1))/__
#define debug(_) cout<<endl<<"d::"<<_<<endl<<endl
#define type(_) typeid(_).name()
inline LL read() {
	LL x = 0, f = 1;char c = getchar();
	while (!isdigit(c)) { if (c == '-')f = -f;c = getchar(); }
	while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48ll), c = getchar();
	return x * f;
}

LL n;
string s;

int main() {
//	freopen("data.txt","r",stdin);
	n=read();
	Fo(i,1,n) {
		cin>>s;
		LL len = s.size() , ff = 0;
		Fo(j,0,len-1)
			if(s[j]!='a') {
				ff = 1;
				break;
			}
		if(!ff) {
			cout<<"NO"<<endl;
			continue;
		}
		if(len%2) {
			LL flag = 0;
			Fo(j,0,len/2-1)
				if(s[j]!=s[len-1-j]) {
					flag = 1;
					break;
				}
			if(!flag) {
				cout<<"YES"<<endl<<"a"<<s<<endl;
			} else {
				Fo(j,0,len-1)
					if(s[j]!='a') {
						cout<<"YES"<<endl<<s.substr(0,len-j)<<"a"<<s.substr(len-j,len)<<endl;
						break;
					}					
			}
		} else {
			LL flag = 0;
			Fo(j,0,len/2-1)
				if(s[j]!=s[len-1-j]) {
					flag = 1;
					break;
				}
			if(!flag)
				cout<<"YES"<<endl<<"a"<<s<<endl;
			else
				Fo(j,0,len-1)
					if(s[j]!='a') {
						cout<<"YES"<<endl<<s.substr(0,len-j)<<"a"<<s.substr(len-j,len)<<endl;
						break;
					}						
		}
	}
	return 0;
}

标签:cout,712Div2,LL,CF,len,flag,Vu,include,define
来源: https://blog.csdn.net/cls1277/article/details/115442359