poj--1426

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23212		Accepted: 9569		Special Judge

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

解题思路：题目的意思是找到n的非零倍数（ nonzero multiple）m，并且m只有0，1组成，m不超过100位，m可能有多个，只要输出其中一个就可以了。简单的dfs。

代码如下：

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int n,flag;
int a[110][2];
int b[2]={0,1};
int res[110];
void dfs(int ans,int id){
	if(flag)return ;
	if(id>100)return ;
	if(ans%n==0){
		flag=1;
		for(int i=1;i<=id;i++){
			printf("%d",res[i]);
		}
		printf("\n");
		return ;
	}
	for(int i=0;i<2;i++){
        if(!a[id+1][b[i]]){
        	a[id+1][b[i]]=1;
        	res[id+1]=b[i];
        	int m=(ans*10+b[i])%n;
        	dfs(m,id+1);
        	a[id+1][b[i]]=0;
        }
	}
}
int main(){
	while(scanf("%d",&n)&&n){
	    if(n==1){
	    printf("1\n");
	    continue;
	    }
	    flag=0;
	    res[1]=1;
		memset(a,0,sizeof(a));
		a[1][1]=1;
	    dfs(1,1);
	}
	return 0;
}