1~9数字问题

/*
*1~9的9个数字，每个数字只能出现一次，要求这样一个9位整数：其第一位能被1整除，前两位能被2整除，
*前三位能被3整除...依次类推，前9位能被9整除。
*/

//方法1 ---> 381654729
#include<iostream>
#include<vector>
#include <iomanip>
bool Used[10];
vector<long long> v;
void Dfs(int k, long long val)
{
  if (k && ((val%k)!= 0))
  {
    return;
  }
  
  if (9 == k)
  {
    v.push_back(val);
    return;
  }

  for (int i = 1; i <= 9; i++)
  {
    if (!Used[i])
    {
      Used[i] = true;
      Dfs(k+1,val*10+i);
      Used[i] = false;
    }
  }
}

int main()
{
  Dfs(0,0);

  for (int i = 0; i < v.size(); i++)
  {
    cout << setprecision(11)<< v[i] << endl;
  }
  return 0;
}

Python代码来自点击打开链接

#Python:

a   = [1,2,3,4,5,6,7,8,9]
out = []

def verify(out):
    length = len(out)
    if length == 0:
        return True
    total = 0
    for i in range(0, length):
        total += pow(10, length - i - 1) * out[i]
    if total % length == 0:
        return True
    else:
        return False

def dfs(out):
    isOK = verify(out)
    if not isOK:
        return
    if len(out) == len(a):
        print out
        return

    unused = []
    for i in a:
        if i not in out:
            unused.append(i)

    for i in unused:
        out.append(i)          #相当于把当前加入的数入栈
        dfs(out)
        out.remove(out[-1])    #考虑加入的数不满足整除条件，则相当于把刚加进去的数出栈

dfs(out)