Hangzhou Electric OJ 1178 (C ++)

Basic questions. Seeking 1 + 3 + 6 + ... + i (i +1) and / 2, the summation formula of N * (N + 1) * (N + 2) / 6, its logarithmic 10, determining the number after E.

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

int main()
{
	int N;
	while (cin >> N)
	{
		if (N == 0)
			break;
		double sum = 1.0 * N * (N + 1) * (N + 2) / 6;
		int num = log10(sum);
		sum /= pow(10, num);
		cout << fixed << setprecision(2) << sum << "E" << num << endl;
	}
	return 0;
}

Keep up.