51nod 1109 01 膜法二进制

给定一个自然数N，找出一个最小的膜法二进制M满足M除以N的余数为0
例如：N = 4，M = 100

 

bfs + 剪枝

 

#include <cstdio>
#include <cstring>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cstring>
#include <stack>
#define MAX_N 2000005
using namespace std;

int n;
int vised[MAX_N];

struct node
{
	string a;
	int num;
}now,nex;


void bfs(){
	memset(vised,0,sizeof(vised));
	queue<node> s;
	now.a="10";
	now.num=10%n;
	vised[now.num]=1;
	s.push(now);
	now.a="11";
	now.num=11%n;
	vised[now.num]=1;
	s.push(now);
	while(!s.empty()){
		now=s.front();
		if(now.num==0)
			cout<<now.a<<endl;
		s.pop();
		for(int i=0;i<=1;i++){
			if(i==0)
				 nex.a=now.a+'0';
			else nex.a=now.a+'1';
			nex.num=(now.num*10+i)%n;
			if(vised[nex.num]==1)	continue;
			vised[nex.num]=1;
			s.push(nex);
		}
	}

}

int main()
{
    while(cin>>n)
    {
        if(n==1)
        {
            printf("1");
            continue;
        }
        else bfs();
    }
}