(1) set the status array p of 10 elements, and record the occurrence of the numbers 0-9 in 6-digit and 3-digit. All array elements are assigned a value of 1, indicating that the numbers 0-9 have not been used.

(2) for each number of attempts, 3 digits x, 6 digits = x*x, and each digit is taken as the subscript of the array. If the corresponding element is 1, the digit will appear for the first time,

Assign a value of 0 to the corresponding element to indicate that the number has appeared once. Otherwise, if the corresponding element is 0, it means that there are duplicate numbers, ending the attempt.

(3) when 9 elements in the state array p are 0, the solution of the problem is found. However, it is necessary to scan the array p once to determine the solution. To avoid this step,

Set a counter K, record the number of different numbers in the process of taking each digit of X, x*x, and find the solution when k=9.

 1 #include<stdio.h>
 2 int main()
 3 {
 4     long x;
 5     int p[10];
 6     int i,t,k;
 7     int num=0;
 8     int n,f,y,m;
 9     for(x=100;x<=999;x++)
10     {
11         for(i=0;i<=9;i++)
12             p[i]=1;
13             y=x*x; //y Represents 6 digits
14             f=y;   //take y Temporarily stored in f in
15             n=x;   //n Represents 3 digits
16         if(f<=999999)
17         {
18             k=0;
19         for(i=1;i<=6;i++)
20         {
21             t=n%10;
22             n=n/10;
23             m=f%10;
24             f=f/10;
25             if(m==t)
26             {
27             k=0;
28             break;
29             }
30             if(i<4)
31                if(p[t]==1)
32                {
33                 p[t]=0;
34                 k++;
35                }
36              else
37               {
38                k=0;
39                break;
40               }
41               if(p[m]==1)
42               {
43                p[m]=0;
44                k++;
45               }
46               else
47               {      
48                k=0;
49                break;
50               }
51         }
52             if(k==9)
53                 {
54                  num=num+1;
55                  printf("%ld\t%d\n",x,y);
56                 }
57         }
58         else
59             continue;
60     }
61             printf("\n");
62             printf("%d\n",num);
63     return 0;
64 }