[P2841] Luo Valley A * B Problem [High Precision] bfs

Subject to the effect:

Topic links: https://www.luogu.org/problemnew/show/P2841
given a number $A$ , you need to give a minimum number $B$ , such that $A$ and $B$ product contains only 0 and 1.

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Ideas:

All processing precision is not explained.
Obviously binary enumeration violence will timeout. Because it will enumerate enumeration completely unnecessary to count the number of the lot.
If the $x$ and $Y$ two meet number 01 $x\equiv y(mod\ n)$且 $x<y$ , then $x$ and $Y$ end plus a 0 or a 1, these two numbers $v \ N$ also congruence. In this case large numbers $(Y)$ it is completely unnecessary to enumerate the prefix $Y$ numbers do not need to enumerate. Because there is always a little more than its numbers and its congruence, and we ask for is the smallest of the answer.
So you can use $bfs$ to enumerate the number 01, to enumerate 0, then enumeration 1, so as to ensure the number 01 is enumerated in ascending order of enumeration. At the same time use $hash$ to determine the number (the array can be used directly over. $n \ leq 10 ^ 4$ ), the same as if you already have a number and the number of remainder, then this would not have continued to enumerate the number down.
Find the answer on exit $bfs$ , because when we have to ensure that the search of the answer as small as possible.
Then in addition to high-precision single actuarial first answer, the second answer is $years$ .

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Code:

#include <queue>
#include <cstdio>
using namespace std;

const int N=10010,M=200;
int n,a[N];
bool hash[N],flag;

struct node
{
	int a[M+1],p,len;
}ans1,ans2;
queue<node> q;

node xx;
void bfs()
{
	xx.len=1; xx.a[1]=1; xx.p=1; q.push(xx);
	hash[1]=1;
	while (q.size())
	{
		node u=q.front();
		q.pop();
		if (!u.p)
		{
			ans1=ans2=u;
			return;
		}
		for (int i=0;i<=1;i++)
		{
			int p=(u.p*10+i)%n;
			if (!hash[p])
			{
				hash[p]=1;
				node v=u;
				v.len++;
				v.a[v.len]=i;
				v.p=p;
				q.push(v);
			}	
		}
	}
}

int main()
{
	scanf("%d",&n);
	if (n==1) return !printf("1 1\n");
	bfs();
	for (int i=1;i<=ans1.len;i++)
	{
		a[i]=ans1.a[i]/n;
		ans1.a[i+1]+=ans1.a[i]%n*10;
	}
	int i=1;
	while (!a[i]) i++;
	for (;i<=ans2.len;i++) printf("%d",a[i]);
	putchar(32);
	for (i=1;i<=ans2.len;i++) printf("%d",ans2.a[i]);
	return 0;
}