POJ 1426 Find The Multiple (BFS)

Find The Multiple

 

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 29062 Accepted: 12057 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

 

题意

求任意一个数是n的倍数,且该数的每一位只能是0或1。

 

思路

利用bfs,从最低的1开始枚举,对于每一位的两种情况进行判断,合理即输出。

 

AC 代码

#include <iostream>
#include<stdio.h>
#include<queue>
#include<math.h>
using namespace std;
typedef __int64 LL;

LL bfs(LL n)
{
    queue<LL>sk;
    sk.push(1);
    while(!sk.empty())
    {
        LL s=sk.front();
        sk.pop();
        if(s%n==0)return s;
        sk.push(s*10);
        sk.push(s*10+1);
    }
    return 0;
}
int main()
{
    LL n;
    while(~scanf("%I64d",&n)&&n)
        printf("%I64d\n",bfs(n));
}