PTA 1012 数字分类 python

PTA 1012 数字分类 (20 分)

输入样例1：

13 1 2 3 4 5 6 7 8 9 10 20 16 18

输出样例1：

30 11 2 9.7 9

输入样例2：

8 1 2 4 5 6 7 9 16

输出样例2：

N 11 2 N 9

Python解法

n = input()
num = list(map(int ,n.split()))
A1 = []
A2 = []
A3 = []
A4 = []
A5 = []
# 统计满足条件的值，存入到数组中
for i in num[1:]:
    if i%5 == 0 and i%2 == 0:
        A1.append(i)
    if i%5 == 1:
        A2.append(i)
    if i%5 == 2:
        A3.append(i)
    if i%5 == 3:
        A4.append(i)
    if i%5 == 4:
        A5.append(i)
# print(A1,A2,A3,A4,A5)
s = 0
s1 = 0
s2 = 0
s2_av = 0
# 判断是否有否和条件的值，这里比较麻烦，也可以用数组做
if len(A1) > 0:
    for i in range(0, len(A1)):
        s += A1[i]
    print(s,end = " ")
else:
    print('N',end = " ")
if len(A2) > 0:
    for i in range(0, len(A2)):
        if i % 2 == 1:
            s1 -= A2[i]
            # print(s1)
        else:
            s1 += A2[i]
            # print(s1)
    print(s1,end = " ")
else:
    print('N', end=" ")
if len(A3) > 0:
    print(len(A3),end = " ")
else:
    print('N',end = " ")
if len(A4) > 0:
    for i in range(0, len(A4)):
        s2 += A4[i]
    s2_av = round(s2 / len(A4), 1)
    print(s2_av,end = " ")
else:
    print('N',end = " ")
if len(A5) > 0:
    A5.sort(reverse=True)
    print(A5[0],end = "")
else:
    print('N',end = "")

如有问题，欢迎在评论区提问。