CodeForces 550C Divisibility by Eight 简单题算是找规律？

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.

Examples

input

3454

output

YES
344

input

10

output

YES
0

input

111111

output

NO

这道题...个人感觉偏向于找规律之类的？反正和网上大牛们的做法不是太一样，大牛们都是找后三位可以被8整除就可以，但是我把表打出来了，可能还是蠢了点

#include <bits/stdc++.h>

using namespace std;

int main()
{
string a;
int len,c[10]={0,6,4,2,0,6,4,2,0,6},d[10]={0,2,0,6,4,2,8,6,0,2};
while(cin>>a)
{
len=a.size();
for(int i=0;i<len;i++)
{
if(a[i]=='0'||a[i]=='8')
{
cout<<"YES"<<endl<<a[i]<<endl;
goto A;
}
else
{
for(int j=i+1;j<len;j++)
{
if(a[j]==c[a[i]-'0']+'0')
{
cout<<"YES"<<endl<<a[i]<<a[j]<<endl;
goto A;
}
}
}
if((a[i]-'0')&1)
{
for(int j=i+1;j<len;j++)
{
if(d[a[j]-'0']!=0)
{
for(int k=j+1;k<len;k++)
{
if(a[k]==d[a[j]-'0']+'0')
{
cout<<"YES"<<endl<<a[i]<<a[j]<<a[k]<<endl;
goto A;
}
}
}
}
}
}
cout<<"NO"<<endl;
A:;
}
return 0;
}