NOJ Palindromes

Palindromes

时间限制(普通/Java) : 1000 MS/ 3000 MS          运行内存限制 : 81920 KByte
总提交 : 65            测试通过 : 36 

题目描述

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

分析：判断输入的字符串是否回文，仅考虑字符串中的字母，我们可以将输入字符串中的字母存在另一个数组里进行判断。

实现代码：

<span style="font-size:12px;">#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int T;int main(){  //  freopen("data.in","r",stdin);    scanf("%d",&T);    getchar();  lp:while(T--)    {        char a[200000],b[200000];        a[0]='\0';        b[0]='\0';        gets(a);        int cnt=0;        for(int i=0;i<strlen(a);i++)        {            if((a[i]>='A'&&a[i]<='Z')||(a[i]>='a'&&a[i]<='z'))                b[cnt++]=a[i];        }        for(int i=0,j=cnt-1;i<=j;i++,j--)        {            if(b[i]!=b[j]&&b[i]!=b[j]+32&&b[i]+32!=b[j])            {                printf("no\n");                goto lp;            }        }        printf("yes\n");    }}</span>