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.
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.
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.
3454
YES 344
10
YES 0
111111
NO
题目大意：就是给你一个字符串，让你去掉其中的几个数，看是否能被8整除
解题思路：就是暴力解决，如果能被8整除，只需要看最后三位数就行
上代码：
/*
Date : 2015-09-04 晚上
Author : ITAKING
Motto :
今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char str[105];
int main()
{
while(cin>>str+2)
{
str[0] = str[1] ='0';
int len = strlen(str), ret;
bool flag = false;
for(int i=0; i<len-2; i++)
{
if(flag)
break;
for(int j=i+1; j<len-1; j++)
{
if(flag)
break;
for(int k=j+1; k<len; k++)
{
ret = (str[i]-'0')*100+(str[j]-'0')*10+str[k]-'0';
if(ret%8 == 0)
{
flag = true;
break;
}
}
}
}
if(flag)
printf("YES\n%d\n",ret);
else
puts("NO");
}
return 0;
}
