[class2] (백준 1259) 팰린드롬수

문제

어떤 단어를 뒤에서부터 읽어도 똑같다면 그 단어를 팰린드롬이라고 한다. 'radar', 'sees'는 팰린드롬이다.

수도 팰린드롬으로 취급할 수 있다. 수의 숫자들을 뒤에서부터 읽어도 같다면 그 수는 팰린드롬수다. 121, 12421 등은 팰린드롬수다. 123, 1231은 뒤에서부터 읽으면 다르므로 팰린드롬수가 아니다. 또한 10도 팰린드롬수가 아닌데, 앞에 무의미한 0이 올 수 있다면 010이 되어 팰린드롬수로 취급할 수도 있지만, 특별히 이번 문제에서는 무의미한 0이 앞에 올 수 없다고 하자.

입력

입력은 여러 개의 테스트 케이스로 이루어져 있으며, 각 줄마다 1 이상 99999 이하의 정수가 주어진다. 입력의 마지막 줄에는 0이 주어지며, 이 줄은 문제에 포함되지 않는다.

출력

각 줄마다 주어진 수가 팰린드롬수면 'yes', 아니면 'no'를 출력한다.

제한

예제 입력 1

121
1231
12421
0

예제 출력 1

yes
no
yes

힌트

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출처

ICPC > Regionals > South Pacific > South Pacific Region > New Zealand Programming Contest > NZPC 2006 B번

• 문제를 번역한 사람: kks227

#include <iostream>
using namespace std;

int Reverse(int n)
{
	int result = 0;
	int powValue = 10000;
	int place = 1;
	bool isPlaceCheck = false;
	for (int i = 0; i < 5; i++)
	{
		if (n / powValue != 0)
		{
			isPlaceCheck = true;
			result += (n / powValue) * place;
			n %= powValue;
		}

		if (isPlaceCheck)
			place *= 10;

		powValue /= 10;
	}
	return result;
}
int main()
{
	int N;
	string result;
	while (1)
	{
		cin >> N;
		if (N == 0)
			return 0;
		result = N == Reverse(N) ? "yes" : "no";
		cout << result << endl;
	}

	return 0;
}

 

 

 

#include <iostream>
using namespace std;

int Reverse(int n)
{
	int result=0;
	while (n > 0)
	{
		result *= 10;
		result += n % 10;
		n /= 10;
	}
	return result;
}
int main()
{
	int N;
	string result;
	while (1)
	{
		cin >> N;
		if (N == 0)
			return 0;
		result = N == Reverse(N) ? "yes" : "no";
		cout << result << endl;
	}

	return 0;
}