Divisibility by Eight

来源：互联网发布：手机吉他软件下载编辑：程序博客网时间：2024/06/05 17:46

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
YES344
input
10
output
YES0
input
111111
output
NO

一个100位以下的数，能否通过删除一些字符使其能被8整除。听说有人打表过，因为被8整除的数只需要后三位被8整除即可。因为大于1000的部分一定可以用8的倍数表示。

因此这个数是否被8整除取决于后三位。一共124个，打表找。我想可以一个一个找，一共才100位，并不会超时，一开始也可以通过找0和8来剪枝。然后分2位和3位来找。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char c[105];
int leap1[105], leap2[105], leap3, pre, sum;
int main()
{
int i, j, m, n, len, leap, k;
scanf("%s",c);
memset(leap1, 0, sizeof(leap1));
memset(leap2, 0, sizeof(leap2));
len = strlen(c); leap = 0; leap3 = 0;
for (i = 0; i < len; i++)
{
  if (c[i] == '0')
  {
   leap = 1;
  }
}
if (leap == 1)
  cout << "YES" << endl << "0" << endl;
else
{
  for (i = 0; i < len; i++)
  {
   if (c[i] == '8')
   {
    leap = 1;
   }
  }
  if (leap == 1)
   cout << "YES" << endl << "8" << endl;
  else
  {
   for (i = 0; i < len; i++)
   {
    for (j = i + 1; j < len; j++)
    {
     if (((c[i] - '0') * 10 + (c[j] - '0')) % 8 == 0)
     {
      cout << "YES" << endl << c[i] << c[j] << endl;
      leap = 1;
      break;
     }
    }
    if (leap == 1)break;
   }
   if (!leap){
    for (i = 0; i < len; i++)
    {
     for (j = i+1; j < len; j++)
     {
      for (k = j+1; k < len; k++)
      {
       if (((c[i] - '0') * 100 + (c[j] - '0') * 10 + (c[k] - '0')) % 8 == 0)
       {
        cout << "YES" << endl << c[i] << c[j] << c[k] << endl;
        leap = 1;
        break;
       }
      }
      if (leap == 1)break;
     }
     if (leap == 1)break;
    }
   }
   if (leap == 0)cout << "NO" << endl;
  }
}
return 0;
}

0 0