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122A. Lucky Division ​

brute force, number theory, 1000, https://codeforces.com/problemset/problem/122/A

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number n is almost lucky.

Input

The single line contains an integer n (1 ≤ n ≤ 1000) — the number that needs to be checked.

Output

In the only line print "YES" (without the quotes), if number n is almost lucky. Otherwise, print "NO" (without the quotes).

Examples

input

47

output

YES

input

16

output

YES

input

78

output

NO

Note

Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.

In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.

python
n = int(input())

for i in {4,7,47,74,447,474,477,747,774}:
    if n%i == 0:
        print('YES')
        break
else:
    print('NO')

2020fall-cs101-赵春源

Python
def check(x):
    s = str(x)
    for c in s:
        if c!= '4' and c!= '7':
            return False
    return True

n = int(input())
for i in range(1, n+1):
    if n%i == 0 and check(i) == True:
        print('YES')
        break
else:
    print('NO')
python
def check(x):
    s = str(x)
    return [False, True][s.count('4') + s.count('7') ==len(s)]

n = int(input())
for i in range(1, n+1):
    if n%i == 0 and check(i) == True:
        print('YES')
        break
else:
    print('NO')

用len函数是多么简单好用，学到了~

Python
n = int(input())
luckynumbers = []
for i in range(1, 1002):
    s = str(i)
    if s.count('4') + s.count('7') == len(s):
        luckynumbers.append(i)

b = 0
for i in luckynumbers:
    if n%i == 0:
        b = 1
print(['NO', 'YES'][b])

any(x)判断x对象是否为空对象，如果都为空、0、false，则返回false，如果不都为空、0、false，则返回true

all(x)如果all(x)参数x对象的所有元素不为0、''、False或者x为空对象，则返回True，否则返回False

python
n = int(input())
print('NO' if all([n%i for i in (4,7,47,74,447,474,477,747,774)]) else 'YES')

2022fall-cs101，刘子芊。分解因数法。

python
import re
#import sys
n = int(input())

#求因数
def factor(n):
    fac = []
    for i in range(1, (n+1)//2):
        if n%i == 0:
            fac.append(i)
            fac.append(n//i)
    return list(set(fac))

#分解因数+本身（无1）
facn = factor(n)

#因数中若出现lucky number则是almost lucky
pattern = r'^[47]+$'
for num in facn:
    if re.match(pattern, str(num)) != None:
        print('YES')
        #sys.exit()
        break
else:
    print('NO')