XJTU Summer Holiday Test 1(Divisibility by Eight-8的倍数)

最新推荐文章于 2019-10-22 13:32:40 发布

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探讨如何从一个不超过100位的非负整数中移除部分数字，使其保留至少一位数字且结果能被8整除，同时保持原有数字顺序不变。通过检查最后三位数字是否为8的倍数来快速判断。

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C - Divisibility by Eight

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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 Input

Input

Output

YES
344

Input

Output

YES
0

Input

Output

NO

8*125=1000 故一个数是8的倍数当且仅当它末尾3个数是8的倍数

所以答案最多有3位，暴力即可

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
char s[MAXN];
int main()
{
//	freopen("I.in","r",stdin);
//	freopen(".out","w",stdout);
	
	scanf("%s",s);
	char *pch=strstr(s,"AB");
	if (pch!=NULL&&strstr(pch+2,"BA")!=NULL)
	{
		cout<<"YES"<<endl;
		return 0;
	}
	pch=strstr(s,"BA");
	if (pch!=NULL&&strstr(pch+2,"AB")!=NULL)
	{
		cout<<"YES"<<endl;
		return 0;
	}
	cout<<"NO"<<endl;
	
	return 0;
}