[HOJ 10136] Palindromes_palindromes ud-优快云博客

本文链接：https://blog.youkuaiyun.com/chenhanxuan1999/article/details/111566671

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。回文是指正读反读都一样的字符序列。文章提供了样例输入输出，并展示了一种通过忽略大小写和非字母字符来检查字符串是否为回文的有效方法。

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Palindromes

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

Total submit users: 1853, Accepted users: 1481

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

AC:

#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>   
#include <vector>

std::vector <char> v;

char t(int& ch)
{
    return ch >= 'A' && ch <= 'Z' ? ch -= ('A' - 'a') : ch;
}

int main()
{
    // freopen("C:/Users/kiko/Desktop/test.txt", "r", stdin);
    int ch;
    while (ch = getchar())
    {
        if (ch != '\n' && ch != -1) v.push_back(t(ch));
        else
        {
            for (int i = 0; i < v.size() / 2; ++i) if (v[i] != v[v.size() - 1 - i]) v.clear();
            if (!v.empty()) { printf("yes\n"); v.clear(); }
            else printf("no\n");
            if (ch == -1) break;
        }     
    }
    return 0;
}