CCF 201912-1 report (score 100)

CCF 201912-1 reports

#include<iostream>

using namespace std;

int main()
{
	int n;
	cin >> n;
	int a = 0 , b = 0 , c = 0 , d = 0;
	// 四个数分别存放甲乙丙丁中各自跳过的数之和
	int count;
	for (int i = 1,count=1; count < n+1; i++,count++)
	{
		if (i % 4 == 1)
		{
			if (i % 7 == 0 || (i%10==7 || i/10%10==7 || i / 100 % 10 == 7 || i / 1000 % 10 == 7)) {
				a++;
				count--;
			}
		}
		else if (i % 4 == 2)
		{
			if (i % 7 == 0 || (i % 10 == 7 || i / 10 % 10 == 7 || i / 100 % 10 == 7 || i / 1000 % 10 == 7)) {
				b++;
				count--;
			}
		}
		else if (i % 4 == 3)
		{
			if (i % 7 == 0 || (i % 10 == 7 || i / 10 % 10 == 7 || i / 100 % 10 == 7 || i / 1000 % 10 == 7)) {
				c++;
				count--;
			}
		}
		else
		{
			if (i % 7 == 0 || (i % 10 == 7 || i / 10 % 10 == 7 || i / 100 % 10 == 7 || i / 1000 % 10 == 7)) {
				d++;
				count--;
			}
		}
	}
	cout << a << endl;
	cout << b << endl;
	cout << c << endl;
	cout << d << endl;
	
	return 0;
}