[UVa]10699:Count the factors

直接從2開始到根號N去除除看能不能整除，
能整除就知道其質因數有此數，因此就把質因數個數加一，
接著把N中所有含有的這個質因數除乾淨，再往下一個搜尋。

[C++](0.012)

	#include<iostream>
	#include<cmath>
	using namespace std;
	int main()
	{
	int n;
	while( cin >> n && n != 0 )
	{
	int output = n;
	int factor = 0;
	int limit = static_cast<int>(sqrt(n));
	for( int i = 2 ; i <= limit ; i++ )
	if( n % i == 0 )
	{
	factor++;
	while( n % i == 0 ) n /= i;
	}
	if( n != 1 )
	factor++;
	cout << output << " : " << factor << endl;
	}
	return 0;
	}

view raw Zerojudged120.cpp hosted with ❤ by GitHub