题解 | #KiKi判断上三角矩阵#
KiKi判断上三角矩阵
https://www.nowcoder.com/practice/9a6786c28cdb45f9b991685f867b5d08
n=int(input())
list1=[]
for i in range(n):
a=list(map(int,input().split()))
list1.append(a)
num=0
for j in range(1,n):
for k in range(j):
if list1[j][k]==0:
num+=1
else:
break
if num==n*(n-1)//2:
print("YES")
else:
print("NO")
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