poj1426Find The Multiple(bfs ohohoh ～～～～好害怕)

最新推荐文章于 2018-12-21 15:38:33 发布

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本文介绍了一个编程问题，即给定一个正整数n，找出一个只包含0和1的非零倍数m。通过广度优先搜索算法解决该问题，并提供了一个C++实现示例。

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26508		Accepted: 10977		Special Judge

Description

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

Input

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

Output

For each value of n in the input(投入) print a line containing the corresponding value of m. The decimal(小数) representation(代表) of m must not contain more than 100 digits(数字). If there are multiple solutions(解决方案) for a given value of n, any one of them is acceptable(可接受的).

Sample Input

Sample Output

10
100100100100100100
111111111111111111

Source

#include<stdio.h>
#include<queue>
using namespace std;
int n;
long long int x;
void bfs()
{
    queue<long long int>q;
    q.push(1);

    while(!q.empty())
    {
      x=q.front();
      q.pop();
      if(x%n==0)
      {
          printf("%lld\n",x);
          break;
      }

      q.push(x*10);
      q.push(x*10+1);
    }
}
int main()
{

   while(~scanf("%d",&n),n)
   {
       bfs();
   }


}