南邮 OJ 1384 Palindromes

Palindromes

时间限制(普通/Java) : 1000 MS/ 3000 MS          运行内存限制 : 81920 KByte
总提交 : 69            测试通过 : 40 

比赛描述

Write a program that determines if each input string is a palindrome. A palindrome is a

string that reads exactly the same in both forward and reverse directions. For something

to be considered a palindrome, it must be at least 1 character long. For the purposes of

your program, ignore any characters that are not letters, as well as spaces when

determining if a string is a palindrome.

输入

The first line of input contains an integer N that indicates the number of test strings to

follow. On each subsequent line there will be a single test string. Here is a sample:

输出

For each test string, output "yes" if the string was a palindrome, and "no" if it was not a

palindrome. Remember: Ignore any characters that are not letters, as well as spaces.

样例输入

5
able ##was I, e****re I s.aw $Elba
this is not a palindrome
A man, a plan, a canal, Panama
another random string
Sator Arepo Tenet Opera Rotas

样例输出

yes
no
yes
no
yes

提示

undefined

题目来源

Internet

#include<iostream>#define MAX_N 1000char c1[MAX_N],c2[MAX_N];int main(){int N,i,j,len;scanf("%d",&N);getchar();while(N--){scanf("%[^\n]",c1);getchar();len = (int)strlen(c1);for(i=j=0;i<len;i++){if(c1[i]>='a' && c1[i]<='z'){c2[j++] = c1[i];}else if(c1[i]>='A' && c1[i]<='Z'){c2[j++] = c1[i]+'a'-'A';}}len = j;for(i=0;(i<<1)<len;i++){if(c2[i]!=c2[len-1-i]){break;}}if((i<<1)<len){printf("no\n");}else{printf("yes\n");}}}