POJ - 1426 Find The Multiple （BFS、打表）

https://vjudge.net/problem/POJ-1426

POJ - 1426 Find The Multiple

• • 题目
  • 分析
  • AC代码
  • • BFS
    • 打表

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

分析

直接打表

AC代码

BFS

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
typedef long long LL;
int main()
{
	LL n;
	while(~scanf("%lld",&n) && n!=0)
	{

		if(n==1)
		{
			printf("1\n");
			continue;
		}

		queue<LL> q;
		q.push(1);

		while(q.size())
		{
			LL t1=q.front();
			LL t2;
			q.pop();

			t2=t1*10;
			q.push(t2);
			if(t2%n==0)
			{
				printf("%lld\n",t2);
				break;
			}

			t2=t1*10+1;
			q.push(t2);
			if(t2%n==0)
			{
				printf("%lld\n",t2);
				break;
			}

		}

	}
	return 0;
}

打表

#include<iostream>
#include<cstdio>
using namespace std;
int h[100];
long long fun(int x)
{
	int i=0;
	while((x/2)!=0)
	{
		h[i]=x%2;
		i++;
		x/=2;
	}
	h[i]=x;
	long long sum=0;
	for(int j=i;j>=0;j--)
		sum=sum*10+h[j];
	return sum;
}
int main()
{
	long long n;
	while(~scanf("%lld",&n))
	{
		if(n==0) break;
		for(int i=1;;i++)
		{
			long long res=fun(i);  //对于每个可能的数，都将其先转化为2进制，看其能否被n整除   比如20 4的2进制100刚好能将其整除 
			if(res%n==0)
			{
				printf("%lld\n",res);
				break;
			}
		}
	}
	return 0;
 }