题解 | #KiKi判断上三角矩阵#
KiKi判断上三角矩阵
http://www.nowcoder.com/practice/9a6786c28cdb45f9b991685f867b5d08
n = int(input()) a, flag = [], 0 for i in range(n): b = list(map(int, input().split())) a.append(b) for j in range(1, n): for k in range(j): if a[j][k] != 0: flag = 1 print('NO') break if flag == 1: break if flag != 1: print('YES')