acm hnu 10136 Palindromes

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本文介绍了一个简单的程序设计问题：如何判断一个字符串是否为回文。通过忽略大小写差异，该程序能够正确判断输入的单词是否正读反读都相同。

Palindromes

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 184, Accepted users: 175

Problem 10136 : No special judgement

Problem description

Background:

Palindromes are strings that read the same both forwards and backwards. `Eye' is one such example (ignoring case). In this problem, you get to write a program to determine if a given word is a palindrome or not.

Input

Each line of input contains one word with no embedded spaces. Each word will have only alphabetic characters (either upper or lower case).

Output

For each line of input, output either `yes' if the word is a palindrome or `no' otherwise. Don't print the quotes. Case should be ignored when checking the words.

Sample Input

eyE
laLAlal
Foof
foobar

Sample Output

yes
yes
yes
no

Problem Source

UD Contest

#include <stdio.h>
#define YES 1
#define NO 0
main()
{
      int flag;
      char s[1000];
      int i,j,n;

      while( scanf("%s",s) > 0 ){
             n = 0;
             while( s[n] != '/0' ) n++;
             for( i = 0 ; (i+1) <= (n/2) ; i++ ){
                  if( tolower(s[i])!= tolower(s[n-1-i]) ){flag=NO;break;}
                  else flag=YES;
             }
             if(flag==NO)printf("no/n");
             else printf("yes/n");

      }
return 0;
}