Find The Multiple poj 1426 bfs

寻找特定倍数

最新推荐文章于 2021-08-18 11:10:35 发布

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本文介绍了一个算法挑战，即为给定的整数n找到一个仅包含0和1的非零倍数m。文章提供了几种不同的实现方法，包括使用队列进行深度优先搜索和递归方法。

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 28309		Accepted: 11735		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

Sample Output

10
100100100100100100
111111111111111111

#include <iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;


long long dfs(int n)
{
   queue<long long>q;
   q.push(1);
   while(1)
   {
      long long c=q.front();
      q.pop();
      if(c%n==0)
      return c;
      q.push(c*10+1);
      q.push(c*10);
   }
}
int main()


{
   int n;
   while(cin>>n&&n)
   {
      cout<<dfs(n)<<endl;
   }
    return 0;
}
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int dfs(long long pre,long long x,long long n){
if(pre>x) return 0;//判断是否溢出，溢出了就不是正解
if(x%n==0)
{
  cout<<x<<endl;return 1;
}
if(dfs(x,x*10,n)) return 1;
return dfs(x,x*10+1,n);
}
int main()
{
   int n;
   while(scanf("%d",&n)!=EOF&&n)
   {
       dfs(0,1,n);
   }
}
#include <iostream>
#include <stdio.h>
using namespace std;


typedef unsigned long long LL;  //这里可以用long long


int f;


LL ans;


void dfs(LL n,LL mu,int s)
{
    if(f||(s==19))
        return;
    if(mu%n==0)
    {
        f=1;
        ans=mu;
        return;
    }
    else
    {
        dfs(n,mu*10,s+1);
        dfs(n,mu*10+1,s+1);
    }
    return;
}
int main()
{
    int n;
    while(1)
    {
        f=0;
        scanf("%d",&n);
        if(!n)
            break;
        LL nn=n;
        dfs(nn,1,0);
        printf("%I64u\n",ans);
    }
    return 0;
}