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POJ Find The Multiple 1426 (搜索)

时间:2015-10-22 23:47:26      阅读:228      评论:0      收藏:0      [点我收藏+]

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Find The Multiple
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 22576   Accepted: 9291   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
题目大意:找到n的一个倍数,并且这个数只由0 1组成。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 int n;
 7 bool flag;
 8 unsigned long long  m;
 9 void dfs(unsigned long long  x,int k)
10 {
11     cout<<x<<endl;
12     if(flag)
13         return;
14     if(x%n==0)
15     {
16         flag=1;
17         m=x;
18         return;
19     }
20     if(k==19)
21         return;
22     dfs(x*10,++k);
23     dfs(x*10+1,k);
24     return;
25 }
26 int main()
27 {
28     int i,j;
29     while(cin>>n&&n)
30     {    
31         flag=0;
32         dfs(1,1);
33         cout<<m<<endl;
34     }
35 }

 

POJ Find The Multiple 1426 (搜索)

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原文地址:http://www.cnblogs.com/a1225234/p/4903086.html

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