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): 6289    Accepted Submission(s): 1458

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 37 8 9100 10

 

Sample Output

Case 1: 2345Case 2: -1

 

Source

2012 Asia Chengdu Regional Contest

 

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听说这是个数位bfs，就是用bfs去暴力数的每一位，这里要用到同余定理，既余数只要相同，后面加一个数字，在取余的时候所得到的余数也相同，这样找到最后余数为0即可，用个pre数组记录下路径，然后就可以了

My ugly code

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <iostream>#include <queue>using namespace std;const int maxn=1e4+10;int n,m;int a[maxn];bool del[10],vis[maxn];/*del用于记录哪个数据可以用,vis记录余数，出现过的就不再重复出现*/int pre[maxn];/*记录前驱，方便输出*/char text[maxn];/*记录走过的值*/int cas=1;bool bfs(){    queue<int> q;    while(!q.empty())        q.pop();    memset(pre,-1,sizeof(pre));    memset(vis,false,sizeof(vis));    q.push(0);    int cur;    while(!q.empty()){        cur=q.front();        q.pop();        for(int i=0;i<=9;i++){            if(del[i] || cur==0&&i==0)                continue ;            int yu=(cur*10+i)%n;            if(vis[yu])                continue ;            vis[yu]=true;            q.push(yu);            text[yu]='0'+i;            pre[yu]=cur;            if(yu == 0){                return true;            }        }    }    return false;}void out(){    char pout[maxn];    int cnt=0;    int cur=0;    pout[cnt++]=text[cur];    cur=pre[cur];    while(cur != 0){        pout[cnt++]=text[cur];        cur=pre[cur];    }    pout[cnt]='\0';    printf("%s\n",_strrev(pout));}int main(){    while(~scanf("%d%d",&n,&m)){        int tmp;        memset(del,false,sizeof(del));        for(int i=1;i<=m;i++){            scanf("%d",&tmp);            del[tmp]=true;        }        bool ans=bfs();        printf("Case %d: ",cas++);        if(ans)            out();        else            printf("-1\n");    }    return 0;}