Codeforces Round #313 (Div. 2) D. Equivalent Strings
D. Equivalent Strings
Time Limit: 2 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/559/problem/B
Description
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are calledequivalent in one of the two cases:
- They are equal.
- If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
- a1 is equivalent to b1, and a2 is equivalent to b2
- a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Sample Input
aaba
abaa
Sample Output
YES
aabb
abab
NO
HINT
题意
判断俩字符串是否类似就是说一半的一半是否相同,交换的情况下又是否相同
题解:
暴力
代码
#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <string>
#include <stack>
#include <math.h>
#include <vector>
#include <string.h>
using namespace std; typedef __int64 ll;
const int INF = (int)1E9+;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
} //*******************************
bool isEqual(char* a,char* b,int len)
{
bool flag=true;
for(int i=;i<len&&flag;i++)
{
if(a[i]!=b[i]) flag=false;
}
if(flag) return true;
if(len&) return false;
return (isEqual(a,b,len/)&&isEqual(a+len/,b+len/,len/))||(isEqual(a+len/,b,len/)&&isEqual(a,b+len/,len/));
} int main()
{
char a[];
char b[];
scanf("%s%s",a,b);
int len=strlen(a);
if(len%)
{
if(strcmp(a,b)==)
{
printf("YES\n");
}
else printf("NO\n");
return ;
}
if(isEqual(a,b,len))
{
printf("YES\n");
}
else printf("NO\n"); return ;
}
Codeforces Round #313 (Div. 2) D. Equivalent Strings的更多相关文章
- Codeforces Round #313 (Div. 1) B. Equivalent Strings
Equivalent Strings Problem's Link: http://codeforces.com/contest/559/problem/B Mean: 给定两个等长串s1,s2,判断 ...
- Codeforces Round #313 (Div. 1) B. Equivalent Strings DFS暴力
B. Equivalent Strings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559 ...
- Codeforces Round #313 (Div. 2) 560D Equivalent Strings(dos)
D. Equivalent Strings time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- Codeforces Round #313 (Div. 2) D.Equivalent Strings (字符串)
感觉题意不太好懂 = =# 给两个字符串 问是否等价等价的定义(满足其中一个条件):1.两个字符串相等 2.字符串均分成两个子串,子串分别等价 因为超时加了ok函数剪枝,93ms过的. #includ ...
- Codeforces Round #313 (Div. 2)(A,B,C,D)
A题: 题目地址:Currency System in Geraldion 题意:给出n中货币的面值(每种货币有无数张),要求不能表示出的货币的最小值.若全部面值的都能表示,输出-1. 思路:水题,就 ...
- Codeforces Round #313 (Div. 2) 解题报告
A. Currency System in Geraldion: 题意:有n中不同面额的纸币,问用这些纸币所不能加和到的值的最小值. 思路:显然假设这些纸币的最小钱为1的话,它就能够组成随意面额. 假 ...
- codeforces 559b//Equivalent Strings// Codeforces Round #313(Div. 1)
题意:定义了字符串的相等,问两串是否相等. 卡了时间,空间,不能新建字符串,否则会卡. #pragma comment(linker,"/STACK:1024000000,102400000 ...
- Codeforces Round #313 (Div. 2) A.B,C,D,E Currency System in Geraldion Gerald is into Art Gerald's Hexagon Equivalent Strings
A题,超级大水题,根据有没有1输出-1和1就行了.我沙茶,把%d写成了%n. B题,也水,两个矩形的长和宽分别加一下,剩下的两个取大的那个,看看是否框得下. C题,其实也很简单,题目保证了小三角形是正 ...
- Codeforces Round #302 (Div. 1) C. Remembering Strings DP
C. Remembering Strings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
随机推荐
- CF360B Levko and Array (二分查找+DP)
链接:CF360B 题目: B. Levko and Array time limit per test 2 seconds memory limit per test 256 megabytes i ...
- 设置dom元素可拖动,支持ie5+
摘要: 最近在项目中要做一个图片预览的功能,这时候会遇到用户上传很大的图片,已经超出视图界面.最终决定做一个在固定宽和高的位置,用户可以拖动图片查看.所以自己就写了一个支持ie5+,chrome,Fi ...
- 聊下并发和Tomcat线程数(Updated)
最近一直在解决线上一个问题,表现是: Tomcat每到凌晨会有一个高峰,峰值的并发达到了3000以上,最后的结果是Tomcat线程池满了,日志看很多请求超过了1s. 服务器性能很好,Tomcat版本是 ...
- 用WP_Query自定义WordPress 主循环
我们知道操作 WordPress 主循环(WordPress Loop)最容易的方法是使用 query_posts 函数. 但是使用 query_posts 直接修改 WordPress 默认的主循环 ...
- ACM Computer Factory(dinic)
ACM Computer Factory Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5596 Accepted: 1 ...
- 压缩js和css, IIS开启Etags, IIS开启Gzip
我们在前端页面性能调优时,经常会压缩js和css,下面列出几个比较好用的在线工具. http://www.jb51.net/tools/jsmin/index.htm http://javascrip ...
- [BZOJ1998][Hnoi2010]Fsk物品调度
[BZOJ1998][Hnoi2010]Fsk物品调度 试题描述 现在找工作不容易,Lostmonkey费了好大劲才得到fsk公司基层流水线操作员的职位.流水线上有n个位置,从0到n-1依次编号,一开 ...
- 将linux用在开发环境中
我是如何将linux用在开发环境中的 1.为什么不直接安装Linux在主机 一直想深入学习一下linux的使用,于是将家里的笔记本装了linux系统,但是要将自己的系统打造一个适合开发的环境确实是一件 ...
- 基于Matlab的MMSE的语音增强算法的研究
本课题隶属于学校的创新性课题研究项目.2012年就已经做完了,今天一并拿来发表. 目录: --基于谱减法的语音信号增强算法..................................... ...
- linux 下如何查看和踢除正在登陆的其它用户 ==>Linux下用于查看系统当前登录用户信息的4种方法
在linux系统中用pkill命令踢出在线登录用户 由于linux服务器允许多用户登录,公司很多人知道密码,工作造成一定的障碍 所以需要有时踢出指定的用户 1/#who 查出当前有那些终端登录(用 ...