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));
}
超时了⌇●﹏●⌇
呜呜，可能后台数据做了增强