백준 6975번 Deficient, Perfect, and Abundant

문제 링크입니다: https://www.acmicpc.net/problem/6975

6975번: Deficient, Perfect, and Abundant

간단한 구현 문제였습니다.

 

	#include <iostream>
	using namespace std;
	void getResult(int num)
	{
	int sum = 0;
	for (int i = 1; i < num; i++)
	{
	if (num%i == 0)
	{
	sum += i;
	}
	}
	cout << num;
	if (sum < num)
	{
	cout << " is a deficient number.\n";
	}
	else if (sum == num)
	{
	cout << " is a perfect number.\n";
	}
	else
	{
	cout << " is an abundant number.\n";
	}
	cout << "\n";
	}
	int main(void)
	{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int N;
	cin >> N;
	for (int n = 0; n < N; n++)
	{
	int num;
	cin >> num;
	getResult(num);
	}
	return 0;
	}

view raw .cpp hosted with ❤ by GitHub

 

개발환경:Visual Studio 2017

 

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