题解 | #杨辉三角的变形#

题目的关键点在于找规律，如下代码可以打印出从第三行开始的奇数和偶数的规律（0表示偶数，1表示奇数）

def func(n):
    odd = [1, 1, 1]
    for i in range(3, n+1):
        odd = [1, odd[0]^odd[1]] + [odd[j]^odd[j+1]^odd[j+2] for j in range(len(odd) - 2)] + [odd[-2]^odd[-1], 1]
        print(odd)
    return odd.index(0) if 0 in odd else -1 
  

"""
func(100)
[1, 0, 1, 0, 1]
[1, 1, 0, 1, 0, 1, 1]
[1, 0, 0, 0, 1, 0, 0, 0, 1]
[1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1]
[1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1]
[1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1]
[1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1]
[1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1]
[1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1]
[1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1]
[1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1]
[1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1]
[1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1]
[1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1]
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
[1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1]
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1]
[1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1]
"""

经过观察发现每4行偶数出现的位置就会循环一次（位置分别是[2, 3, 2, 4])

while True:
    try:
        n = int(input())
        if n == 1 or n == 2:
            print(-1)
            continue
        n -= 3 
        n %= 4 
        res = [2,3,2,4]
        print(res[n])
    except EOFError: break