[kuangbin带你飞]专题一 简单搜索E

E - Find The Multiple

 

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

题意：给你一个数n让你找一个只有0和1组成的数可以被n整除。

思路：简单深搜，直接每次*10或者*10+1就行；有两个坑点：1.Special judge YES。2.因为数会很大，超过long long范围（正负9223372036854775807）的时候就直接return ；

#include<iostream>#include<cmath>#include<queue>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n;bool flag;void dfs(int k,long long int sum)//k为这个数的位数，当这个数的位数超过long long范围（19位）就停止 {if(k==20||flag==true){return ;}if(sum%n==0){flag=true;cout<<sum<<endl;return ;}dfs(k+1,sum*10);dfs(k+1,sum*10+1);}int main(){while(cin>>n&&n){flag=false;dfs(1,1);}}