Rank the number of shards

/* Number of shards in the row:
Tom is staring at the number 203879 in a daze.
Originally, 203879*203879 = 41566646641
What's amazing about this? Observe carefully that 203879 is a six-digit number, and the numbers on each digit are different, and all the digits after the square do not appear to constitute its own number.
There is another 6-digit feature. Please find it!
Next, we will summarize the filtering requirements:
1. 6-digit positive integer
2. The numbers on each digit are different.
3. Each digit of the number of shards does not include any integral digits of the original number.
The answer is a 6-digit positive integer.
Submit the answer in a browser.
Note: Only the other six digits are submitted. do not submit the information already given in the question. */
[Cpp]
# Include "stdio. h"
# Include "stdlib. h"
# Include "time. h"
Int xiangtong (int a [], int k)
{For (int c1 = 0; c1 <k; c1 ++)
For (int c2 = 0; c2 <k; c2 ++)
If (c1! = C2 & a [c1] = a [c2])
Return 1; // indicates the same
Return 0; // different
}
Int main ()
{Long kk; // Save the result after Square
Long int start, finish;
Start = clock ();
For (long int I = 102345; I <987654; I ++) // The long type cannot be used here, and an error will occur.
{Int a [6], B [12], k = 0, k1 = 0; // each bit in I is saved.
Long temp = I;
While (temp> 0)
{A [k ++] = temp % 10;
Temp = temp/10;
} // The low position of a's storage quantity, and the high position of a's storage quantity
If (! Xiangtong (a, k ))
{Kk = I * I;
While (kk> 0)
{B [k1 ++] = kk % 10;
Kk = kk/10 ;}
Int key = 0; // The value ranges from 1 to 0.
For (int c1 = 0; c1 <6; c1 ++)
For (int c2 = 0; c2 <k1; c2 ++)
If (a [c1] = B [c2]) {key = 1; break ;}
If (! Key)
{Printf ("% lld ^ 2 = % lld \ n", I, I * I );}
} // End if
} // End
Finish = clock ();
Printf ("\ nall time is: % lfs", (finish-start)/1000.0 );
Printf ("\ n ");
System ("pause ");
}

# Include "stdio. h"
# Include "stdlib. h"
# Include "time. h"
Int xiangtong (int a [], int k)
{For (int c1 = 0; c1 <k; c1 ++)
For (int c2 = 0; c2 <k; c2 ++)
If (c1! = C2 & a [c1] = a [c2])
Return 1; // indicates the same
Return 0; // different
}
Int main ()
{Long kk; // Save the result after Square
Long int start, finish;
Start = clock ();
For (long int I = 102345; I <987654; I ++) // The long type cannot be used here, and an error will occur.
{Int a [6], B [12], k = 0, k1 = 0; // each bit in I is saved.
Long temp = I;
While (temp> 0)
{A [k ++] = temp % 10;
Temp = temp/10;
} // The low position of a's storage quantity, and the high position of a's storage quantity
If (! Xiangtong (a, k ))
{Kk = I * I;
While (kk> 0)
{B [k1 ++] = kk % 10;
Kk = kk/10 ;}
Int key = 0; // The value ranges from 1 to 0.
For (int c1 = 0; c1 <6; c1 ++)
For (int c2 = 0; c2 <k1; c2 ++)
If (a [c1] = B [c2]) {key = 1; break ;}
If (! Key)
{Printf ("% lld ^ 2 = % lld \ n", I, I * I );}
} // End if
} // End
Finish = clock ();
Printf ("\ nall time is: % lfs", (finish-start)/1000.0 );
Printf ("\ n ");
System ("pause ");
}