POJ 1426 Find The Multiple（kuangbin搜索专题）

 

Find The Multiple

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42281		Accepted: 17760		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

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

提示：该题意思就是让你找到个只由0和1组成的数并且这个数可以整除n，可以用深搜（DFS）来完成。

　　　找只由0和1组成的数可以从1开始，递推×10和×10+1，一直到找到该数为止。

 

代码实现如下（g++）:

#include <cstdio>
#define ll long long

using namespace std;

int m;
int flag;

void DFS(ll num,int k)
{
    if(k==19)//如果到达19位，就结束
    {
        return ;
    }
    if(flag)//如果找到该数
    {
        return ;
    }
    if(num%m==0)
    {
        flag=1;
        printf("%lld\n",num);
        return ;
    }
    DFS(num*10,k+1);//深搜num*10
    DFS(num*10+1,k+1);//深搜num*10+1
}


int main()
{
    while(~scanf("%d",&m)&&m)
    {
        flag=0;//不要忘记赋初值
        DFS(1,0);//从1开始搜
    }
    return 0;
}