将1到9重新排列，使得前n位数字组…

最终结果是：381654729；

首先，偶数位上一定是偶数，因为只有偶数才可以被偶数整除；

编程实现如下：

 

#include <iostream>
usingnamespace std;
void main()
{
  int a[9];
  int j, num;
  a[4] =5;
  cout<<"所求的九位数为："<<endl;
  for(a[0] = 1; a[0] < 10; a[0] += 2)
  {
    for (a[1] =2; a[1] < 9;a[1] += 2)
    {
     for (a[2] = 1; a[2] < 10; a[2] += 2)
     {
       if (a[2] == a[0])
         continue;
       for (a[3] = 2; a[3] < 9; a[3] += 2)
       {
         if (a[3] == a[1])
           continue;
         for (a[5] = 2; a[5] < 9; a[5] += 2)
         {
           if (a[5] == a[3] || a[5] == a[1])
             continue;
           for (a[6] = 1; a[6] < 10; a[6] += 2)
           {
             if (a[6] == a[4] || a[6] == a[2] || a[6] == a[0])
               continue;
             for (a[7] = 2; a[7] < 9; a[7] += 2)
             {
               if (a[7] == a[5] || a[7] == a[3] || a[7] == a[1])
                 continue;
               for (a[8] = 1; a[8] < 10; a[8] += 2)
               {
                 if (a[8] == a[6] || a[8] == a[4] || a[8] == a[2] || a[8] == a[0])
                   continue;
                 num = 0;
                 for (j = 0; j< 9; j++)
                 {
                   num = num * 10 + a[j];
                   if(num % (j + 1) != 0)
                     break;
                 }
                 if (j == 9)
                   cout<<num<<endl;
               }
             }
           }
         }
       }
     }
    }
  }
}