Lover to meet the problem. Java code

Problem/*
* 8 Valentine Questions:
*
* Problem Description:
* Place 8 lovers in a 8x8 chessboard, asking each Valentine 22 not to conflict
* (in each row, vertical column, oblique column only one lover).
*
* Data indicates:
* Use a 8-bit 8-digit number to indicate the position of the lover on the chessboard:
* For example: 45615353 means:
* No. 0 Lover in 4th place
* 1th lover in 5th place
* 2nd lover in 6th place
* 。。。
* 7th lover in 3rd place
*
* The process of looping variables from 00000000 to 77777777 (8 numbers) traverses all the cases of the lover
* The program uses octal numbers in a one-dimensional array data[] to represent
*
* Detect conflicts:
* Row conflict: data[i] = = Data[j]
* Diagonal column conflict: (data[i]+i) = = (Data[j]+j) or (data[i]-i) = = (DATA[J]-J)
*
Benefits
* Using loops rather than recursion, with less system resources
* can calculate n Valentine's problem
* The problem of linearization processing, you can block the problem, in a distributed environment with multiple computers together.
*
* ToDo:
* The enumeration section can also be optimized, and more judgment conditions can be faster.
* Output part can be modified into a checkerboard form of output
*
* @author CinC 2002-09-11
*
*/

public class Queen {
int size;
int resultcount;

public void compute (int size) {
this.size = size;
Resultcount = 0;
int data[] = new Int[size];
int count; Number of all possible cases
int i,j;

Calculate the number of all possible cases
Count = 1;
for (i=0; i<size; i++) {
Count = count * size;
}
For every possible situation
for (i=0; i<count; i++) {
Calculates the position of the lover on the chessboard in this case, using a 8-digit representation
Here you can optimize
int temp = i;
for (j=0; j<size; j) {
data [j] = temp% size;
temp = temp/size;
}
Test the feasibility of this situation, if possible, output
if (Test (data))
Output (data);
}
}

/*
* Test this situation if the lover's arrangement is feasible
*
*/
public boolean test (int[] data) {
int i,j;
for (i=0; i<size; i++) {
for (j=i+1; j<size; j) {
Test whether in the same row
if (data[i] = = Data[j])
return false;
Test whether in a slash
if ((data[i]+i) = = (DATA[J]+J)
return false;
Test whether in a backslash
if ((data[i]-i) = = (DATA[J]-J)
return false;
}
}
return true;
}

/*
* Output the coordinates of a lover in some case
*
*/
public void output (int[] data) {
int i;
System.out.print (++resultcount + ":");
for (i=0; i<size; i++) {
System.out.print ("+ i +", "+ data[i] +" ");
}
System.out.println ();
}

The main () is here.
public static void Main (String args[]) {
(New Queen ()). Compute (8);
}
}