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
给出一个数n,找出一个数要求是n的倍数,并且这个数的十进制只由1和0组成, 两条搜索路径—t*10;t*10+1
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
bool vis;
int n;
void dfs(unsigned long long x,int n,int k)
{
if(vis)
return;
if(x%n==0)
{
cout<<x<<endl;
vis=true;
return;
}
if(k==19)
return;
dfs(x*10,n,k+1);
dfs(x*10+1,n,k+1);
}
int main(){
while(scanf("%d",&n))
{
if(n==0) break;
vis=false;
dfs(1,n,0);
}
return 0;
}
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