s = 0
for i in range(1,2021):
    s += str(i).count('2')
print(s)
# 624

nums = [[2, 2, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0],
        [0, 0, 2, 2, 0, 2],
        [0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 2, 2],
        [0, 0, 2, 0, 2, 0],
        ]
n = 0
for i in range(len(nums)):
    for j in range(len(nums[0])):
        if j+4 <= len(nums[0]):
            a = nums[i][j:j+4]
            if a == [2,0,2,0]:
                n+=1
        if i+3<len(nums):
            b = nums[i][j]*1000+nums[i+1][j]*100+nums[i+2][j]*10+nums[i+3][j]
            if b == 2020:
                n+=1
        if i+3<len(nums) and j+3 < len(nums[0]):
            c = nums[i][j]*1000+nums[i+1][j+1]*100+nums[i+2][j+2]*10+nums[i+3][j+3]
            if c == 2020:
                n+=1
print(n)

import datetime
a = datetime.date(2000,1,1)
b = datetime.date(2020,10,2)
num = 0
while True:
    if a == b:
        break
    if a.day == 1:
        num+=2
        a = a+datetime.timedelta(days=1)
        continue
    if a.isoweekday() == 1:
        num+=2
        a = a + datetime.timedelta(days=1)
        continue
    num+=1
    a = a + datetime.timedelta(days=1)
print(num)
# 8879

# 代码运行
lst = [[] for i in range(40)]
num = 1
lst[0].append(1)
for i in range(2, 40):
    if i % 2 == 0:
        for j in range(i):
            num += 1
            lst[j].append(num)
    else:
        for j in range(i):
            num += 1
            lst[i - j - 1].append(num)
for i in range(len(lst)):
    print(i + 1, lst[i])
print(lst[19][19])

# 当所有的字符全全部倒序的时候重新排序的次数就越多，所用的字符就越少
# 又因为要求字典序最小，所以字母应该是前n位
# 由于冒泡排序的特征，完全倒序的情况下，交换的次数是(n-1)+(n-2)......2+1
# 算出当有15字符完全倒序的时候为，105，所以，只需要在此基础上稍加修改即可，
# 又因为需要保证字典序最小，即字符串的第一个字目应该尽量的小，所以将j前提5个字符
# 'abcdefghijklmno'
# print('jonmlkihgfedcba')

n = int(input())
y = 0
j = 0
for i in range(n):
    num = int(input())
    if num >= 85:
        y+=1
    if num >=60:
        j +=1
print('{:.0f}%'.format((j/n)*100))
print('{:.0f}%'.format((y/n)*100))