View Code of Problem 31

#include <iostream>
#include<cmath>
using namespace std;
int main() {
	int n;
	while (cin>>n)
	{
		int num = 0;
		for (int i = 9; ; i=i+7)//找到最小的符合条件的值
		{
			if (i % 3 == 2 &&i%5==3) {
				num=i;
				break;
			}	
		}
		cout << num + 105 * (n - 1) << endl;//加上3 5 7 的最小公倍数的倍数即可
	}
	
}