题目:
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
题意:
本题要找出数m,m是只有0和1构成的十进制数,并且是n的倍数,若有多个答案,输出任意一个就可以。
题解:
经过思考会发现m最大不会超过unsigned long long 的范围,所以用unsigned long long保存就可以,接下来就是深搜就ok。
代码:
#include <iostream>
using namespace std;
unsigned long long ans;
bool f;
void dfs(unsigned long long s,int n,int k)
{
if(f) return ;
if(s%n==0) {ans=s;f=true; return ;}
if(k==19) return ; //如果k超过19就不在unsigned long long的范围内了
dfs(s*10,n,k+1);
dfs(s*10+1,n,k+1);
return ;
}
int main()
{
int n;
while(cin>>n,n)
{
ans=0;
f=false;
dfs(1,n,0);
cout<<ans<<endl;
}
return 0;
}