POJ1426-Find The Multiple(BFS||DFS)

Posted 2020-09-06

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

Means:

给你一个数n要你找出能整除n的一个只有0||1组成的十进制数,随便输出一个都行

Solve:

水题,其实就是从1开始枚举每次为p * 10 || p * 10 + 1,然后,蜜汁爆CPP,蜜汁ACGPP

Code:

 1 #include <queue>
 2 #include <cstdio>
 3 using namespace std;
 4 typedef unsigned long long ll;
 5 ll n;
 6 inline void Bfs()
 7 {
 8     queue<ll> qu;
 9     qu.push(1);
10     while(!qu.empty())
11     {
12         ll p = qu.front();
13         qu.pop();
14         if(p % n == 0)  {printf("%lld\n" , p); return ;}
15         qu.push(p * 10);
16         qu.push(p * 10 + 1);
17     }
18 }
19 int main()
20 {
21     while(~scanf("%lld" , &n) && n)
22     {
23         Bfs();
24     }
25 }

View Code