POJ 1426 Find The Multiple

POJ 1426 Find The Multiple

题目链接http://poj.org/problem?id=1426

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

题意：

给你一个数字n然后找出由0，1组成的数字可以整除这个数字。

题解：

可以使用查找，首先第一位一定是1.然后对于后面有两种可能，一种是加0，一种是加1.可以使用队列，同时不断取余。对于一个数字来说（ab）%mod == (a%mod)(b%mod);(a+b)%mod == (a%mod) + (b%mod);

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
long long solve(int n)
{
	queue <long long> qu;
	qu.push(1);
	long long temp;
	while (!qu.empty()){
		temp = qu.front();
		qu.pop();
		if ((temp*10)%n == 0)
			return temp*10;
		if ((temp*10+1)%n == 0)
			return temp*10+1;
		qu.push(temp*10);
		qu.push(temp*10+1);
	}
	return -1;
}
int main(int argc, char const *argv[])
{
	int n;
	while (scanf("%d",&n) && n){
		long long ans = solve(n);
		cout<<ans<<endl;
	}
	return 0;
}