Find The Multiple poj 1426 bfs

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 28309 Accepted: 11735 Special Judge

Description

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

26190

Sample Output

10100100100100100100111111111111111111

#include <iostream>#include<queue>#include<cstdio>#include<cstring>using namespace std;long long dfs(int n){   queue<long long>q;   q.push(1);   while(1)   {      long long c=q.front();      q.pop();      if(c%n==0)      return c;      q.push(c*10+1);      q.push(c*10);   }}int main(){   int n;   while(cin>>n&&n)   {      cout<<dfs(n)<<endl;   }    return 0;}#include<cstdio>#include<cstring>#include<iostream>using namespace std;int dfs(long long pre,long long x,long long n){if(pre>x) return 0;//判断是否溢出，溢出了就不是正解if(x%n==0){  cout<<x<<endl;return 1;}if(dfs(x,x*10,n)) return 1;return dfs(x,x*10+1,n);}int main(){   int n;   while(scanf("%d",&n)!=EOF&&n)   {       dfs(0,1,n);   }}#include <iostream>#include <stdio.h>using namespace std;typedef unsigned long long LL;  //这里可以用long longint f;LL ans;void dfs(LL n,LL mu,int s){    if(f||(s==19))        return;    if(mu%n==0)    {        f=1;        ans=mu;        return;    }    else    {        dfs(n,mu*10,s+1);        dfs(n,mu*10+1,s+1);    }    return;}int main(){    int n;    while(1)    {        f=0;        scanf("%d",&n);        if(!n)            break;        LL nn=n;        dfs(nn,1,0);        printf("%I64u\n",ans);    }    return 0;}