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): 2370 Accepted Submission(s): 545

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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Statistic | Submit | Discuss | Note

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include <map>
 3 #include <queue>
 4 #include <vector>
 5 #include <string>
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <cstdlib>
10 #include <iostream>
11 #include <algorithm>
12 using namespace std;
13 #define maxn 200005
14 #define ll long long
15 #define INF 0x7fffffff
16 #define eps 1e-8
17 int n, m;
18 queue<int>q;
19 int pre[maxn];
20 int rear[maxn];
21 int a[maxn];
22 void dfs(int u){
23     if (pre[u] != -1)dfs(pre[u]);
24     printf("%d", rear[u]);
25 }
26 void bfs(){
27     int u;
28     while (!q.empty())q.pop();
29     for (int i = 1; i < 10; i++)if (!a[i]){
30         if ( (i % n == 0)){ printf("%d", i); return; }
31         else if(!pre[i%n]){ pre[i%n] = -1; rear[i%n] = i; q.push(i%n); }
32     }
33     int t = 0;
34     while (!q.empty()&&t<=n){
35         t++;
36         u = q.front(); q.pop();
37         for (int i = 0; i < 10; i++)if (!a[i]){
38             int v;
39             v = (u * 10 + i) % n;
40             if (!pre[v]){
41                 pre[v] = u; rear[v] = i;
42                 if (v == 0){ dfs(v); return; }
43                 q.push(v);
44             }
45         }
46     }
47     printf("-1");
48 }
49 int main(){
50     int cas = 1;
51     int t;
52     while (~scanf("%d%d",&n,&m)){
53         memset(a, 0, sizeof a);
54         memset(pre, 0, sizeof pre);
55         for (int i = 0; i < m; i++){ scanf("%d", &t); a[t] = 1; }
56         printf("Case %d: ", cas++);
57         bfs();
58         puts("");
59     }
60     return 0;
61 }

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posted @ 2013-10-29 03:20 HaibaraAi 阅读(95) 评论(0) 收藏举报

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