Thinking analysis: 9 -bit integers, consisting of 1 ~ 9 without repeated, pay attention to analyzing without 0.

Each number only appears once, so the smallest number is 123456789, and the largest number is 987654321.

Since it is not repeated, then the 9 numbers can be compared with each one. How to compare, 9 numbers are used in A! = B && A! = C …

Want to put the obtained numbers in the array, and compare the number of numbers by bubbling sorting method. Don’t say that bubbling sort can only be used for sorting …

improvement code can be used.

is not 0 and the numbers are different. After the two conditions are met, they continue to judge.

Use a cycle to judge that the first two are divided by 2, and the top three are divided by 3. …, the top nine are removed by 9. Conditions meet the output and end.

Answer: 381654729

Detailed code is as follows:

```
public class testlargenumber {
/**
* @param ARGS
*/
Public Static Void Main (String [] args) {{
// Todo Auto-Generatd Method Stub
int [] wei = new int [9]; //
int Flag;
int i, num, cNT = 0, n, SUM = 0; // The number of statistics is not equal and 0
for (i = 123456789; I <= 987654321; i ++) {
num = i; // Initialize num, if I use i directly, it will change later, be careful!
Flag = 0; // Used to determine whether to meet the conditions
// Calculate each bit
for (int j = 0; j <9; j ++) {
wei [j] = num % 10;
// system.out.println (wei [j]);
num = num / 10;
}
// Judgment is 0 or equal
for (int k = 0; k <9; k ++) {
if (wei [k] == 0) {{
Flag = 1;
Break;
}
for (int m = k+1; m <9; m ++) {
if (wei [k] == wei [m]) {{
Flag = 1;
break; // Jump out of the inner cycle
}
}
if (flag == 1) {
BREAK; // Jump out of the outer cycle
}
}
// Satisfy none of 0 and each person is not equal, and continue to judge that the first few of them have been removed
if (flag == 0) {
cNT = 0; // count to 0 to 0
SUM = Wei [8];
for (n = 8; n> = 1; n-) {
SUM = SUM*10+Wei [n-1];
if (SUM%(10-N) == 0) {{
cNT ++; // Increase
}
}
if (cNT == 8) {// The first 2 digits are divided into the top 9 and 9 are divided into 8.
System.out.println (i); // I used i here, and used num to execute
}
}
}
// system.out.println (cNT);
}
}
```