poj 1426 Find The Multiple

题目链接：点击打开链接

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

题目大意：给出一个n　找出是n的倍数m　而且m是只有0和1组成的

基本思路：dfs和bfs都可以，我用的bfs;首先用long long 存储是可以放开的，然后就是暴力搜索，从1开始然后对其乘１０和乘１０＋１，直到出现结果．

我看的别人的博客都用同的dfs 一开始写的bfs也是超内存，但是想到用队列可以去除队前的数，从而使得数据会减少从而ac

<span style="font-size:18px;">#include <iostream>#include<cstring>#include<queue>using namespace std;unsigned long long b;queue<unsigned long long>a;int main(){    int n;    while(cin>>n,n)    {        while(!a.empty())        {            a.pop();        }        a.push(1);        while(!a.empty())        {            b=a.front();            a.pop();            if(b%n==0)            {                break;            }            else            {                a.push(b*10);                a.push(b*10+1);            }        }        cout<<b<<endl;    }    return 0;}</span>