Problem--41A--Codeforces--Translation

本文介绍了一个简单的编程问题:如何判断一个字符串是否是另一个字符串的逆序。通过C语言实现了一个小算法,该算法接受两个字符串作为输入,并确定第二个字符串是否为第一个字符串的逆序。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A. Translation
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it’s easy to make a mistake during the «translation». Vasya translated word s from Berlandish into Birlandish as t. Help him: find out if he translated the word correctly.

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

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

Examples
input
code
edoc
output
YES
input
abb
aba
output
NO
input
code
code
output
NO

01

#include<stdio.h>
#include<string.h>
int main(){
    char str1[105],str2[105],count=0;
    scanf("%s%s",str1,str2);
    int length=strlen(str1);
    for(int i=0;i<length;i++)
    {
        if(str1[i]==str2[length-i-1])
            count++;
    }
    count==length?printf("YES\n"):printf("NO\n");
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值