Run ID:67060

提交时间:2024-03-13 20:22:31

""" 4 1 0 1 1 0 0 0 0 1 1 1 1 0 1 0 1 如果矩阵符合条件,则输出OK;如果不符合,输出Corrupt """ n = int(input()) a = [] for x in range(n): tmp = input().split() a.append(tmp) flag = True for row in a: if row.count('1')%2==1: flag = False break if flag: for y in range(n): cnt =0 for row in range(n): if a[row][y]=='1': cnt+=1 if cnt%2==1: flag = False break if flag==True: print("OK") else: print("Corrupt")