POJ 1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21832 Accepted: 8975 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

题目传送门：POJ1426 Find The Multiple

题目挺简单的，就是输入一个数n，让你求一个数能整除n。但是这个数只能由0和1组成。

思路就是BFS来破咯，设需要求的这个数为x。x从1开始，如果能整除n就输出，不能的话。分别求出x=x*10，x=x*10+1，也就是在x后填了1或0，再去看能不能整除。还不能整除就再重复上述操作。挺简单的吧~  题目上面写着 Special Judge，说明答案不止一个，输出一个符合题目要求的就ok啦~ 下面是AC代码【本题交G++，交C++超时】

#include <cstdio>#include <iostream>#include <queue>using namespace std;void bfs(int n){    queue<long long>Q;    Q.push(1);    while(!Q.empty())    {        long long x;        x=Q.front();        Q.pop();        if(x%n==0)        {            printf("%lld\n",x);            return ;        }        Q.push(x*10);        Q.push(x*10+1);    }}int main(){    int n;    while(scanf("%d",&n) &&n)    {        bfs(n);    }    return 0;}