Divisibility by Eight

最新推荐文章于 2022-07-07 10:34:57 发布

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本文探讨了如何通过删除一个不超过100位的非负整数中的某些数字，使得剩余数字组成的数能够被8整除的问题。文章提供了一种算法实现思路，包括特殊情况的快速判断方法以及遍历寻找符合条件子串的过程。

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

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;
}