poj 1426(同余搜索)

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <string>
#include <iostream>
using namespace std;
int n;
bool mod[210]; ///这里是最强的剪枝,同余剪枝,如果一个小数的余数和大数的余数相同
               ///假设小点的数 a ,大点的为 b ,a%n==b%n => (a+c)%n == (b+c)%n ,我们假设 a+c是符合条件的,那么
               /// b+c 也是符合条件的,所以我们可以直接不要 b+c 了,这就是一个很巧妙并且很强的剪枝.
struct Node{
    int mod;
    string ans;
};
string bfs(){
    memset(mod,false,sizeof(mod));
    queue<Node> q;
    Node s;
    s.ans = "1";
    s.mod = 1%n;
    q.push(s);
    while(!q.empty()){
        Node now = q.front();
        q.pop();
        if(now.mod==0){
            return now.ans;
        }
        Node next;
        next.mod = (now.mod*10+1)%n;
        next.ans = now.ans+"1";
        if(!mod[next.mod]){
            mod[next.mod] = true;
            q.push(next);
        }
        next.mod = (now.mod*10)%n;
        next.ans = now.ans+"0";
        if(!mod[next.mod]){
            mod[next.mod] = true;
            q.push(next);
        }
    }
}
int main(){

    while(scanf("%d",&n)!=EOF,n){
        cout<<bfs()<<endl;
    }
}