poj 1426 BFS
Find The Multiple
Time Limit: 1000MS
Memory Limit:10000K
Total Submissions: 10604
Accepted: 4373
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
1: #include <iostream>
2: //#include <queue>
3: //#include <cstring>
4: #include <cstdio>
5: using namespace std;
6: long long num[9999999];
7:
8: int main()
9: {
10: int n,i,j;
11: while(scanf("%d",&n) && n!=0)
12: {
13: i=0;
14: j=1;
15: num[0] = 1;
16: while(1)
17: {
18: if(num[i]%n == 0)
19: {
20: printf("%I64d\n",num[i]);
21: break;
22: }
23: else
24: {
25: num[j++] = num[i]*10;
26: num[j++] = num[i]*10+1;
27: i++;
28: }
29: }
30: }
31: //system("pause");
32: return 0;//
33: }
34:
35:
这道题目的BFS比较简单，主要需要注意的是long long的利用，队列都不需要用上。在计算时如果直接利用取模运算应该可以进一步降低运行时间，利用二叉树的性质进行回溯即可
