POJ 1426 Find The Multiple(BFS)

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22009 Accepted: 9051 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

10100100100100100100
111111111111111111
#include<iostream>//此题的答案不维一#include<cstdio>#include<cstring>//竟然找一个数的十进制的整数倍还可以这么做#include<cmath>#include<queue>#define LL long longusing namespace std;LL n,m;void bfs(LL x){    LL y;    queue<LL>q;    while(!q.empty())    q.pop();    q.push(x);    while(!q.empty())    {        x=q.front();        q.pop();        if(x%n==0)        {            printf("%lld\n",x);            return ;        }        for(LL i=0;i<2;i++)        {            y=x;            if(i==0)            {                y=y*10+1;                q.push(y);            }            else            {                y=y*10;                q.push(y);            }        }    }}int main(){    while(~scanf("%lld",&n))    {        if(n==0) break;        bfs(1);    }    return 0;}