CodeForces - 550C 思路水题

最新推荐文章于 2019-08-13 14:32:54 发布

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本文介绍了一种算法，用于从一个不超过100位的非负整数中找到可以被8整除的最长子序列。通过检查所有可能的一位、两位和三位数组合，该算法确定是否存在满足条件的子序列，并输出结果。

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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
3454
Output
YES
344
Input
10
Output
YES
0
Input
111111强调内容
Output
NO

题目大意:
给你一个长度小于100的数字，让你删掉其中的一部分或者不删除数字，然剩下的数字可以被8整除。
题目思路
由于1000可以被8整除，所以任何大于1000的数值可以分解为1000*a+b的形式，所以我们只需要关心b即可，这样的话，我们在数字中找1位，2位，3位的数字观察其能否被8整除即可。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;

int main()
{
    string s;
    cin>>s;
    int len=s.length();
    int num=-1;
    for(int i=0;i<len;i++)
    {
        if(s[i]-'0'==8||s[i]-'0'==0)
        {
            num=s[i]-'0'; break;
        }
    }
    for(int i=0;i<len-1;i++)
    {
        for(int j=i+1;j<len;j++)
        {
            int temp=(s[i]-'0')*10+s[j]-'0';
            if(temp%8==0)
            {
                num=temp; break;
            }
        }
        if(num!=-1) break;
    }
    for(int i=0;i<len-2;i++)
    {
        for(int j=i+1;j<len-1;j++)
        {
            for(int k=j+1;k<len;k++)
            {
                int temp=(s[i]-'0')*100+(s[j]-'0')*10+s[k]-'0';
                if(temp%8==0)
                {
                    num=temp; break;
                }
            }
            if(num!=-1) break;
        }
        if(num!=-1) break;
    }
    if(num==-1) printf("NO\n");
    else        printf("YES\n%d\n",num);
    return 0;
}