cf306 C. Divisibility by Eight(数学推导）

C. Divisibility by Eight

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a non-negative integer n, its decimal representation consists of at most 100 digits and doesn't contain leading zeroes.

Your task is to determine if it is possible in this case to remove some of the digits (possibly not remove any digit at all) so that the result contains at least one digit, forms a non-negative integer, doesn't have leading zeroes and is divisible by 8. After the removing, it is forbidden to rearrange the digits.

If a solution exists, you should print it.

Input

The single line of the input contains a non-negative integer n. The representation of number n doesn't contain any leading zeroes and its length doesn't exceed 100 digits.

Output

Print "NO" (without quotes), if there is no such way to remove some digits from number n.

Otherwise, print "YES" in the first line and the resulting number after removing digits from number n in the second line. The printed number must be divisible by 8.

If there are multiple possible answers, you may print any of them.

Sample test(s)

input

3454

output

YES
344

input

10

output

YES
0

input

111111

output

NO

 解题思路：

可被2整除 : 偶數。
可被3整除 : 各位數字之和能被3整除。
可被4整除 : 最末兩位數字所成的2位數能被4整除。
可被5整除 : 末位為 0 或 5。
可被6整除 : 同時滿足被2和3整除的條件。

可被7整除 : 
設一數為 abc def , 則 def - abc 能被7整除 ,
又如 abc def ghi , 則 ghi - def + abc 能被7整除 ,
再如 abc def ghi jkl , 則 jkl - ghi + def - abc 能被7整除, 如此類推。
以上任何字母包括首位可為0 , 此法亦適合判別能否被11或13整除。

可被8整除 : 最末三位數字所成的3位數能被8整除。
可被9整除 : 各位數字之和能被9整除。
可被10整除 :  末位為 0。
可被11整除 :  奇位數字和 -  偶位數字和能被11整除。
可被12整除 :  同時滿足被3和4整除的條件。
可被13整除 :  (同可被7整除之法則)    

 

转载于:https://www.cnblogs.com/get-an-AC-everyday/p/4554053.html