Palindromes+string+reverse()函数

最新推荐文章于 2019-10-25 16:11:27 发布

李逝水

最新推荐文章于 2019-10-25 16:11:27 发布

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分类专栏：字符串文章标签：字符串 String reverse

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本博客介绍了一个简单的程序，用于判断给定的字符串是否为回文串。程序通过直接使用库函数将字符串逆序并比较来实现这一功能。

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Palindromes

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2317 Accepted Submission(s): 1375

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
/*思路：直接用reverse()库函数将字符串逆序，判断是否相等*/
 
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{
	string a;
	int count=1;
	while(cin>>a&&a!="STOP"){
		cout<<"#"<<count++<<": ";
		string b=a;
		reverse(b.begin(),b.end());
		if(a==b)
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}