Yet Another Multiple Problem HDU 4474 BFS

Yet Another Multiple Problem

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2004    Accepted Submission(s): 482

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 ≤ 10⁴). 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

 

Source

2012 Asia Chengdu Regional Contest

 

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#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<iostream>
#include<vector>
#include<queue>
//ios_base::sync_with_stdio(false);
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define sz(v) ((int)(v).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define clr(x) memset(x,0,sizeof(x))
#define clrs( x , y ) memset(x,y,sizeof(x))
#define out(x) printf(#x" %d
", x)
#define sqr(x) ((x) * (x))
typedef long long LL;

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 30000;

int sgn(const double &x) {  return (x > eps) - (x < -eps); }

int n,m;
int cant[maxn];
int vis[maxn];
struct Node
{
    int mod;
    int pre;
    int val;
    int id;
}node[maxn];

void print_ans(int i)
{
    if(node[i].pre != -1)
        print_ans(node[i].pre);
    //cout<<i<<endl;
    printf("%d",node[i].val);
}
void bfs()
{
    clr(vis);
    queue<Node> q;
    int cnt = 0;
    rep(i,1,10)
    {
        if(!cant[i] && !vis[i%n])
        {
            vis[i%n] = 1;
            node[cnt].mod = i%n;
            node[cnt].pre = -1; 
            node[cnt].val = i;
            node[cnt].id = cnt;
            q.push(node[cnt]);
            cnt++;
        }
    }
    while(!q.empty())
    {
        Node temp = q.front();
        q.pop();
        if(temp.mod == 0)
        {
            print_ans(temp.id);
            return ;
        }
        
        int ret;
        rep(i,0,10)
        {
            if(cant[i])
                continue;
            ret  = temp.mod*10 + i;
            int a = ret%n;
            if(!vis[a])
            {
                vis[a] = 1;
                Node t;
                t.mod = a;
                t.val = i;
                t.pre = temp.id;
                t.id = cnt;
                node[cnt] = t;
                cnt++;
                q.push(t);
            }
        }
    }
    printf("-1");
}
int main() 
{
    //freopen("in.txt","r",stdin);
    int k = 1;
    while(scanf("%d%d",&n,&m) == 2)
    {
        clr(cant);
        rep(i,0,m)
        {
            int a;
            scanf("%d",&a);
            cant[a] = 1;
        }
        printf("Case %d: ",k++);
        bfs();
        cout<<endl;
    }
    return 0;
}