A - Translation字符串匹配问题

最新推荐文章于 2021-10-04 20:18:59 发布

原创最新推荐文章于 2021-10-04 20:18:59 发布 · 570 阅读

0 ·

CC 4.0 BY-SA版权

ACM 专栏收录该内容

56 篇文章

订阅专栏

本文探讨了从Berland语言到Birland语言的翻译难题，这两种语言非常相似，只是意义相同的单词拼写和发音相反。文章通过实例展示了如何帮助Vasya检查他是否正确地将单词s翻译成了t。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A -Translation _p

Time Limit: 2000MS

Memory Limit: 262144K

64bit IO Format: %I64d& %I64u

[Submit] [GoBack] [Status]

Description

The translation from the Berland language into the Birlandlanguage is not an easy task. Those languages are very similar: aberlandish word differs from a birlandish word with the samemeaning a little: it is spelled (and pronounced) reversely. Forexample, a Berlandish word code corresponds to aBirlandish word edoc. However, it's easy to make amistake during the «translation». Vasya translatedword s from Berlandishinto Birlandish as t. Help him: find out if hetranslated the word correctly.

Input

The first line contains word s, the secondline contains word t. The words consist of lowercaseLatin letters. The input data do not consist unnecessary spaces.The words are not empty and their lengths do not exceed 100symbols.

Output

If the word t is aword s, written reversely,print YES,otherwise print NO.

Sample Input

Input

code

edoc

Output

YES

Input

abb

aba

Output

NO

Input

code

code

Output

NO

解法一：普通匹配

#include<iostream>
using namespace std;
int main()
{
        char a[101],b[101],c[101];
        int k,count;
        while(scanf("%s %s",a,b)!=EOF)
        {   
                count=0;
                k=strlen(a);
            strcpy(c,strrev(b));
                for(int i=0;i<k;i++)
                {
                        if(c[i]==a[i])
                                count++;
                }
                if(count==k) printf("YES\n");
                else printf("NO\n");
        }
        return 0;

解法二：利用STL中的倒置函数 strrev

#include<iostream>
using namespace std;
int main()
{
      char str2[102],str1[102];
          while(cin>>str1>>str2)
          {
             if(strcmp(str1,strrev(str2))==0)
                         cout<<"YES"<<endl;
                 else
                         cout<<"NO"<<endl; 
          }
  return 0;