搜索-H

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

26190

Sample Output

10100100100100100100111111111111111111

1、题意:给你一个数n,然后求一个能被n整除的数,而且该数只能由0或者1组成.
2、思路:广搜即可.
3、代码:

#include<iostream>#include<queue>#include<stdio.h>using namespace std;void find(int n){queue<__int64> find;find.push(1);while(find.empty()==0){    __int64 num;num=find.front();    find.pop();    if(num%n==0){        printf("%I64d\n",num);            return ;        }        find.push(num*10);        find.push(num*10+1);    }}int main(){int n;while(1){cin>>n;if(n==0)break;find(n);}return 0;}

4、总结:在vc6.0上还没法用longlong......,查资料,然后用__int6的时候又没法用cout,又查资料...,虽然这道题不难,但做这道题真是曲折啊.