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Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

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

26190

Sample Output

10100100100100100100111111111111111111

思路就是1       1+10       1+10+1    加出来的两个分别+10    +10+1    以此类推    类似于二叉树

#include <cstdlib>#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <string>#include <algorithm> using namespace std;unsigned long long answer;int dfs(unsigned long long prenum,unsigned long long lev,int n){    if(prenum>=lev)     return 0;       if(lev%n==0)    {answer=lev;return 1;}        if(dfs(lev,lev*10,n))    {return 1;}         return dfs(lev,lev*10+1,n);     }int main(){    int n;    int i,j,k;         while(scanf("%d",&n)!=EOF)    {         if(!n)break;         if(dfs(0,1,n))cout<<answer<<endl;    }}