Codeforces 550C - Divisibility by Eight(思维)

最新推荐文章于 2019-06-11 19:54:41 发布

Snakeflute

最新推荐文章于 2019-06-11 19:54:41 发布

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探讨如何从一个非负整数中移除部分数字，使其余部分仍为非负整数且能被8整除。通过检查长度为1、2、3的数字片段来寻找解决方案。

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C. Divisibility by Eight
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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

题意:
给出一串数,去除任意多个数,使得剩下的数能被8整除.

解题思路:
1000能被8整除,那么对于一个大于等于1000的数可以分解成1000*n+k.
所以只要看长度为1,2,3的数能否被8整除即可.

AC代码:

#include <bits/stdc++.h>
using namespace std;
string a;
int main()
{
    ios::sync_with_stdio(0);
    cin >> a;
    for(int i = 0; a[i]; i++)
    {
        if((a[i]-'0') % 8 == 0)
            {
                cout << "YES" << endl;
                cout << a[i];
                return 0;
            }
    }
    for(int i = 0; a[i]; i++)
    {
        for(int j = i+1; a[j]; j++)
        {
            if(((a[i]-'0')*10 + a[j]-'0') % 8 == 0)
            {
                cout << "YES" << endl;
                cout << a[i] << a[j];
                return 0;
            }
        }
    }
    for(int i = 0; a[i]; i++)
    {
        for(int j = i+1; a[j]; j++)
        {
            for(int k = j+1; a[k]; k++)
            {
                if(((a[i]-'0')*100 + (a[j]-'0')*10 + a[k]-'0') % 8 == 0)
                {
                    cout << "YES" << endl;
                    cout << a[i] << a[j] << a[k];
                    return 0;
                }
            }
        }
    }
    cout << "NO";
    return 0;
}