HDOJ 题目4474 Yet Another Multiple Problem（bfs，技巧，找一个数的倍数）

最新推荐文章于 2019-08-26 16:18:33 发布

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最新推荐文章于 2019-08-26 16:18:33 发布

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本文介绍了一道关于寻找特定正整数倍数的问题——YetAnotherMultipleProblem，并提供了一个高效的解决方案。该问题要求找到最小的正整数倍数，其十进制表示中不包含给定的一组数字。

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Yet Another Multiple Problem

Time Limit: 40000/20000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4057 Accepted Submission(s): 947

Problem Description

There are tons of problems about integer multiples. Despite the fact that the topic is not original, the content is highly challenging. That’s why we call it “Yet Another Multiple Problem”.
In this problem, you’re asked to solve the following question: Given a positive integer n and m decimal digits, what is the minimal positive multiple of n whose decimal notation does not contain any of the given digits?

Input

There are several test cases.
For each test case, there are two lines. The first line contains two integers n and m (1 ≤ n ≤ 10 ⁴). The second line contains m decimal digits separated by spaces.
Input is terminated by EOF.

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) while Y is the minimal multiple satisfying the above-mentioned conditions or “-1” (without quotation marks) in case there does not exist such a multiple.

Sample Input

Sample Output

  
   Case 1: 2345
Case 2: -1

Source

2012 Asia Chengdu Regional Contest

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看的别人代码，，，，很经典啊，，，收藏一下

ac代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
int v[10100],fin[10100],pre[10100],ans;
char s[10010];
int n,m;
int bfs()
{
	int i,j;
	queue<int>q;
	q.push(0);
	while(!q.empty())
	{
		int cur;
		cur=q.front();
		q.pop();
		for(i=0;i<=9;i++)
		{
			if(fin[i]||i==0&&cur==0)
				continue;
			int mod=(cur*10+i)%n;
			if(!v[mod])
			{
				v[mod]=1;
				s[mod]=i+'0';
				pre[mod]=cur;
				q.push(mod);
				if(mod==0)
				{
				//	ans=cur*10+i;
					return 1;
				}
			}
		}
	}
	return 0;
}
void print()
{
	string ans;
	int p=0;
	while(p!=0||ans.empty())
	{
		ans+=s[p];
		p=pre[p];
	}
	reverse(ans.begin(),ans.end());
	puts(ans.c_str());
}
int main()
{
	//int n,m;
	int c=0;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(v,0,sizeof(v));
		memset(fin,0,sizeof(fin));
		while(m--)
		{
			int num;
			scanf("%d",&num);
			fin[num]=1;
		}
		printf("Case %d: ",++c);
		if(bfs())
		{
			print();
		//	printf("%d\n",ans);
		}
		else
			printf("-1\n");
	}
}