第一位能被1整除，前两位能被2整除，...，前九位能被9整除

Posted 2020-12-26 wqpkita

#include<iostream>
#include<vector>
using namespace std;

void nineBitInteger(int k, long long a, bool *check, vector<long long> &result)
{   //判断是否可以被位数（k）整除，不可则结束
	if (k&&a%k != 0)return;
	//第9位也符合条件则将数A保存起来
	if (k == 9)
	{
		result.push_back(a);
		return;
	}
	//递归过程
	for (int i = 1; i <= 9; i++)
	{
		if (!check[i])
		{  //1-9这九个数，其中检测过则赋值为1
			check[i] = 1;
			nineBitInteger(k + 1, a * 10 + i, check, result);
			check[i] = 0;
		}
	}
}

void main()
{
	bool check[10];
	memset(check, 0, sizeof(check));
	vector<long long> result;
	nineBitInteger(0, 0, check, result);
	for (unsigned int i = 0; i < result.size(); i++)
	{
		cout << result[i] << endl;
	}
	cout << endl;
}