Waterloo Cup--the seventh annual final: Anti-magic square

Anti-Magic Square


China's ancient books are very early records


2 9 4
7 5 3
6 1 8


This is a third-order magic square. Each row and the number on the diagonal are added equally.


Consider one of the opposite questions below.
You can not fill Hu Ci sudoku with 1~9 numbers.
So that the numbers on each diagonal of each row and column are not equal.




This should be done.
Like what:
9 1 2
8 4 3
7 5 6


Your task is to search all third-order anti-magic squares. And count out how many species there are.
Rotate or mirror the same species.


Like what:
9 1 2
8 4 3
7 5 6


7 8 9
5 4 1
6 3 2


2 1 9
3 4 8
6 5 7


And so on are counted in the same situation.

Please submit the third-order anti-magic square altogether how many kinds. This is an integer and do not fill in any excess content.

Direct violence Search bar, the most practical game

public class anti-magic square {static int arr[]={1,2,3,4,5,6,7,8,9};
	static int count=0;
		public static void Main (string[] args) {dfs (0);
	System.out.println (COUNT/8); public static void Dfs (int k) {if (k==arr.length) {if (Arr[0]+arr[1]+arr[2]!=arr[3]+arr[4]+arr[5]&&arr[0]+ar r[1]+arr[2]!=arr[6]+arr[7]+arr[8]&&arr[0]+arr[1]+arr[2]!=arr[3]+arr[0]+arr[6]&&arr[0]+arr[1]+ ARR[2]!=ARR[1]+ARR[4]+ARR[7]&&ARR[0]+ARR[1]+ARR[2]!=ARR[5]+ARR[2]+ARR[8]&&ARR[0]+ARR[1]+ARR[2] !=ARR[0]+ARR[4]+ARR[8]&&ARR[0]+ARR[1]+ARR[2]!=ARR[2]+ARR[4]+ARR[6]) if (Arr[3]+arr[4]+arr[5]!=arr[6]+arr [7]+arr[8]&&arr[3]+arr[4]+arr[5]!=arr[0]+arr[3]+arr[6]&&arr[3]+arr[4]+arr[5]!=arr[1]+arr[4]+ ARR[7]&&ARR[3]+ARR[4]+ARR[5]!=ARR[2]+ARR[8]+ARR[5]&&ARR[3]+ARR[4]+ARR[5]!=ARR[0]+ARR[4]+ARR[8] &&ARR[3]+ARR[4]+ARR[5]!=ARR[2]+ARR[4]+ARR[6]) if (arr[6]+arr[7]+arr[8]!=arr[0]+arr[3]+arr[6]&& Arr[6]+arr[7]+arr[8]!=arr[1]+arr[4]+arr[7]&&aMp;arr[6]+arr[7]+arr[8]!=arr[2]+arr[8]+arr[5]&&arr[6]+arr[7]+arr[8]!=arr[0]+arr[4]+arr[8]&&arr [6]+arr[7]+arr[8]!=arr[2]+arr[4]+arr[6]) if (arr[0]+arr[3]+arr[6]!=arr[1]+arr[4]+arr[7]&&arr[0]+arr[3]+ ARR[6]!=ARR[2]+ARR[8]+ARR[5]&&ARR[0]+ARR[3]+ARR[6]!=ARR[0]+ARR[4]+ARR[8]&&ARR[0]+ARR[3]+ARR[6] !=ARR[2]+ARR[4]+ARR[6]) if (Arr[1]+arr[4]+arr[7]!=arr[2]+arr[8]+arr[5]&&arr[1]+arr[4]+arr[7]!=arr[0]+arr [4]+arr[8]&&arr[1]+arr[4]+arr[7]!=arr[2]+arr[4]+arr[6]) if (arr[2]+arr[5]+arr[8]!=arr[0]+arr[4]+arr[8]& AMP;&ARR[2]+ARR[5]+ARR[8]!=ARR[2]+ARR[4]+ARR[6]) if (arr[0]+arr[4]+arr[8]!=arr[2]+arr[4]+arr[6]) cou
			nt++;
		Return
			for (int i=k;i<arr.length;i++) {int t=arr[k];
			Arr[k]=arr[i];
			arr[i]=t;
			DFS (K+1);
			T=ARR[K];
			Arr[k]=arr[i];
		arr[i]=t; }
	}
}