hdu 4474 Yet Another Multiple Problem

Yet Another Multiple Problem

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4962    Accepted Submission(s): 1087

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

   
   
   
   

    
    
    
    2345 3
7 8 9
100 1
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 2345
Case 2: -1
   
   
   
   

 

题意：告诉一个数n，以及k个数字，求一个n的整数倍，并且这个倍数中不存在这k个数，求最小的倍数值。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

typedef __int64 ll;

struct Node
{
    int yu;
    int digit;
    int pre;
    int id;
}f[1000009];

queue<int>q;
int use[20];
int vis[1000009];
ll n;
int k;
int cnt;
int ca;
void output(int id)//数字可能很大，递归输出
{
    if(f[id].pre==-1)return;
    output(f[id].pre);
    printf("%d",f[id].digit);
}

int bfs()
{
    f[0].yu=0;
    f[0].pre=-1;
    f[0].id=0;
    q.push(0);
    int t;
    cnt=1;

    while(!q.empty())
    {
        t=q.front();
        q.pop();

        for(int i=0;i<=9;i++)
        {
            if(use[i]) continue;
            if(f[t].yu==0 && i==0) continue;

            int yu=(f[t].yu*10+i)%n;
            if(vis[yu]) continue;

            if(yu==0)
            {
                printf("Case %d: ",ca);
                output(t);
                printf("%d\n",i);
                return 1;
            }
            vis[yu]=1;
            f[cnt].digit=i;
            f[cnt].id=cnt;
            f[cnt].yu=yu;
            f[cnt].pre=t;
            q.push(cnt++);
        }
    }
    return -1;
}


int main()
{
    ca=1;
    while(~scanf("%I64d%d",&n,&k))
    {
        memset(vis,0,sizeof vis);
        memset(use,0,sizeof use);

        for(int i=0;i<k;i++)
        {
            int a;
            scanf("%d",&a);
            use[a]=1;
        }

        while(!q.empty()) q.pop();

        int flag=bfs();

       if(flag==-1)
       {
           printf("Case %d: ",ca);
            puts("-1");
       }
       ca++;
    }
    return 0;
}