Find The Multiple

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 64606

 

Accepted: 26229

 

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

Source

​​Dhaka 2002​​

 

一开始以为是一道数学题，做了以后发现似乎并没有什么特别的规律，所以这里其实还是用的二进制枚举，值得注意的是由于空间范围限制的非常小，所以结构体里面存数字用的是bool形式存的。

1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 const int MAX=21;
12 int n;
13 struct Num{
14     bool num[MAX];
15     int len;
16     int mod;
17 }zt,tt;
18 queue <Num> q;
19 int main(){
20     freopen ("multiple.in","r",stdin);
21     freopen ("multiple.out","w",stdout);
22     int i,j;
23     while (scanf("%d",&n),n!=0){
24         while (!q.empty()) q.pop();
25         zt.len=0;
26         zt.num[++zt.len]=1;zt.mod=1%n;
27         q.push(zt);
28         while (!q.empty()){
29             zt=q.front();
30             if (zt.mod==0) break;
31             q.pop();
32             tt=zt;
33             tt.num[++tt.len]=1;tt.mod=(tt.mod*10+1)%n;
34             q.push(tt);
35             tt=zt;
36             tt.num[++tt.len]=0;tt.mod=(tt.mod*10)%n;
37             q.push(tt);
38         }
39         for (i=1;i<=zt.len;i++)
40             printf("%d",zt.num[i]);
41         printf("\n");
42     }
43     return 0;
44