题解 | #KiKi判断上三角矩阵#

/*
	0的个数为1~n-1的等差数列之和
*/
import java.util.Scanner;
public class Main {
    
    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int count = 0;
        int[][] array = new int[n][n];
        for(int i = 0 ; i < n ; i++) {
            for(int j = 0 ; j < n ; j++) {
                array[i][j] = sc.nextInt();
                if(i > j && array[i][j] == 0) {
                    count++;
                }
            }
        }
        if(count == n*(n-1) / 2) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}