题解 | #杨辉三角的变形#
杨辉三角的变形
https://www.nowcoder.com/practice/8ef655edf42d4e08b44be4d777edbf43
# 从第三行开始 首位偶数的位置表现为2,3,2,4,……并循环
n = int(input())
if n <=2:
print(-1)
else:
tmp = (n-2)%4
if tmp == 1 or tmp == 3:
print(2)
elif tmp == 2:
print(3)
else:
print(4)
# # 第2层开始 12321起 每层0-3位元素必有至少一个偶数
# # 只计算前4列
# n = int(input())
# # 表示层数
# if n < 4:
# if n <= 2:
# print(-1)
# if n == 3:
# print(1)
# else:
# # n = 4 开始 考虑a3
# a1 = n - 1
# a2 = (n-1) * (n) / 2
# a3 = 7
# if n > 4:
# # 第5层开始考虑a3累加
# for i in range(4 + 1, n+1):
# a3 += (i-2) + (i-2)*(i-1)/2
# # 上一层的a1 a2
# # print(a1)
# # print(a2)
# # print(a3)
# if a1 % 2 == 0:
# print(2)
# elif a2 % 2 == 0:
# print(3)
# elif a3 % 2 == 0:
# print(4)
# n = int(input())
# # 表示层数
# tower = [[1]]
# for i in range(n-1):
# tower.append([])
# # 创建n层塔
# for i in range(1, n):
# # 开始填充
# for j in range(i+1):
# # 只用输出一半
# tmp_l = 0
# # 左位
# tmp_m = 0
# # 中位
# tmp_r = 0
# # 右位
# # 填充第i层第j位 从0开始
# # print(f'i:{i} j:{j}')
# tmp = 0
# try:
# # print(1)
# tmp_l = tower[i-1][j-2]
# if j < 2:
# tmp_l = 0
# except:
# # print(2)
# tmp_l = 0
# try:
# tmp_m = tower[i-1][j-1]
# if j < 1:
# tmp_m = 0
# except:
# tmp_m = 0
# try:
# tmp_r = tower[i-1][j]
# if j == i:
# tmp_r = tmp_l
# except:
# tmp_r = 0
# if j == i:
# tmp_r = tmp_l
# tmp = tmp_l + tmp_m + tmp_r
# # print(f'tmp_l :{tmp_l }')
# # print(f'tmp_m :{tmp_m }')
# # print(f'tmp_r :{tmp_r }')
# # print(f'tmp : {tmp}')
# tower[i].append(tmp)
# for i in range(len(tower[n-1])):
# tmp_0 = -1
# for j in range(len(tower[i])):
# if tower[i][j] % 2 == 0:
# tmp_0 = j+1
# break
# print(tmp_0, end=' ')
# print(tower[i])
# tmp_0 = -1
# for i in range(len(tower[n-1])):
# if tower[n-1][i] % 2 == 0:
# tmp_0 = i+1
# break
# print(tmp_0)
