Palindromes

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Write a program to determine whether a word is a palindrome. A palindrome is a sequence of characters that is identical to the string when the characters are placed in reverse order. For example, the following strings are palindromes: “ABCCBA”, “A”, and “AMA”.
The following strings are not palindromes: “HELLO”, “ABAB” and “PPA”.

Input

The input file will consist of up to 100 lines, where each line contains at least 1 and at most 52 characters. Your program should stop processing the input when the input string equals “STOP”. You may assume that input file consists of exclusively uppercase
letters; no lowercase letters, punctuation marks, digits, or whitespace will be included within each word.

Output

A single line of output should be generated for each string. The line should include “#”, followed by the problem number, followed by a colon and a space, followed by the string “YES” or “NO”.

Sample Input

ABCCBA
A
HELLO
ABAB
AMA
ABAB
PPA
STOP

Sample Output

#1: YES
#2: YES
#3: NO
#4: NO
#5: YES
#6: NO
#7: NO

我想shi的心都有了，那天考试连A三次都没过，思路也没啥问题啊！事后突然发现大写的O不知怎么的被写写成小写的了。哎，不然名次就上升好几名了

#include<stdio.h>
#include<string.h>
char a[60];
int main()
{
	int t,len,i,j,m=1,b,c,d;
	while(scanf("%s",a)!=EOF)
	{
		if(!strcmp(a, "STOP")) break;
		printf("#%d: ",m++);
		b=c=d=0;
		len=strlen(a);
		if(len==1)    {printf("YES\n");continue;}
		t=len/2;
		if(len%2==0)
		{
			for(i=0;i<t;i++)
			{
				if(a[i]==a[len-1-i])
				b++;
			}
			if(b==t)  printf("YES");
			else    printf("NO");
		}

		else
		{
			for(i=0;i<=t;i++)
			{
				if(a[i]==a[len-1-i])
				c++;
			}
			if(c-1==t)   printf("YES");
			else   printf("NO");
		}
		printf("\n");
	}
	return 0;
}

HDU 2163 Palindromes