C-POJ-1426 Find The Multiple

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

BFS、DFS都可以

BFS

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long LL;
void BFS(int n){
LL ans=1,b;
queue <LL> a;
a.push(ans);
while(!a.empty()){
b=a.front();
a.pop();
if(b%n==0){
printf("%lld\n",b);
return;
}
a.push(b*10);
a.push(b*10+1);
}
return;
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
if(n==0)
break;
BFS(n);
}
return 0;
}

DFS

#include<stdio.h>
#include<string.h>
#include<stdbool.h>
char ans[1005];
bool flag;
void dfs(int mod,int k,int n){
if(k>20){
ans[k]='\0';
return;
}
if(mod==0){
flag=true;
return;
}
if(!flag){
ans[k]='0';
dfs((mod*10)%n,k+1,n);
}
if(!flag){
ans[k]='1';
dfs((mod*10+1)%n,k+1,n);
}
if(!flag){
ans[k]='\0';
}
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
if(n==0)
break;
ans[0]='1';
flag=false;
dfs(1,1,n);
printf("%s\n",ans);
memset(ans,'\0',sizeof(ans));
}
return 0;
}

这题很简单，没啥说的。