Find The Multiple 二叉树 哈夫曼编码 经典，，，同余方程（(a*b)%n=(a%n*b%n)%n (a+b)%n=(a%n+b%n)%n）...

Problem 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

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大神图片经典   领教啦！！

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 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstdio>
 6 #include<queue>
 7 using namespace std;
 8 int mod[600001],res[600001];
 9 int n,i,j,k;
10 int cnt;
11 int main()
12 {
13    while(scanf("%d",&n)&&n)
14    {
15       mod[1]=1%n;
16       for(i=2;mod[i-1];i++)
17       {
18           mod[i]=(mod[i/2]*10+i%2)%n;//哈夫曼树，模仿双向bfs();
19       }
20       i--;
21       int id=0;
22       while(i)
23       {
24           res[id++]=i%2;//哈夫曼编码
25           i/=2;
26       }
27       for(i=id-1;i>=0;i--)
28         printf("%d",res[i]);
29       printf("\n");
30    }
31    return 0;
32 }

View Code

 

转载于:https://www.cnblogs.com/sdau--codeants/p/3364244.html