codeforces 550C Divisibility by Eight_you are given a non-negative integer n, its decima-优快云博客

本文介绍了一个算法问题，即如何从一个非负整数中移除某些数字，使得剩余数字组成的数能被8整除。文章通过示例说明了如何检查并找到符合条件的数字组合。

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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 
    
      Copy
    
3454

     output 
    
      Copy
    
YES
344

     input 
    
      Copy
    
10

     output 
    
      Copy
    
YES
0

     input 
    
      Copy
    
111111

     output 
    
      Copy
    
NO

我的做法是暴力，判断了最多3位数，当大于3位数时，写错了，错的原因是，单独的一位数，如果它是8或0都可以整除8，我写成了==8，因此在第20组数据--303，错在这组数据了。

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
    // freopen("shuju.txt","r",stdin);
    char a[110];
    cin>>a;
    int len=strlen(a);
    if(len==1)
    {
        if((a[0]-'0')%8==0)
        {
            printf("YES\n");
            printf("%d\n",a[0]-'0');
        }
        else
            printf("NO\n");
    }
    else if(len==2)
    {
        int ans=(a[0]-'0')*10+a[1]-'0';
        if(ans%8==0)
        {
            printf("YES\n");
            printf("%d\n",ans);
        }
        else if((ans/10)%8==0)
        {
            printf("YES\n");
            printf("%d\n",ans/10);
        }
        else if((ans%10)%8==0)
        {
            printf("YES\n");
            printf("%d\n",ans%10);
        }
        else
            printf("NO\n");
    }
    else
    {
        for(int i=0;i<len;i++)
        {
            for(int j=i+1;j<len;j++)
            {
                for(int k=j+1;k<len;k++)
                {
                    int x=a[i]-'0',y=a[j]-'0',z=a[k]-'0';
                    if(x%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",x);
                        return 0;
                    }
                    if(y%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",y);
                        return 0;
                    }
                    if(z%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",z);
                        return 0;
                    }
                    int ans=((x*10)+y)*10+z;
                    if(ans%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",ans);
                        return 0;
                    }
                    ans=x*10+y;
                    if(ans%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",ans);
                        return 0;
                    }
                    ans=x*10+z;
                    if(ans%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",ans);
                        return 0;
                    }
                    ans=y*10+z;
                    if(ans%8==0)
                    {
                        printf("YES\n");
                        printf("%d\n",ans);
                        return 0;
                    }
                }
            }
        }
        printf("NO\n");
    }
    return 0;  
}