POJ1426——BFS——Find The Multiple

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

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

大意：输入一个数，要求输出一个数满足是这个数的倍数，而且只由1和0构成，BFS用c++交超时，g++过了 (*@ο@*)

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

～