Sequence I

Posted 2020-09-17 happy_code

 

Sequence I (hdu 5918)

 

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1938    Accepted Submission(s): 730

Problem Description

Mr. Frog has two sequences a1,a2,⋯,an and b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bmis exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤n and q≥1.

 

 

Input

The first line contains only one integer T≤100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106.

The second line contains n integers a1,a2,⋯,an(1≤ai≤109).

the third line contains m integers b1,b2,⋯,bm(1≤bi≤109).

 

 

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

 

 

Sample Input

2

6 3 1

1 2 3 1 2 3

1 2 3

6

3 2 1

3 2 2 3 1 1

2 3

 

Sample Output

Case #1: 2

Case #2: 1

 

 

//题意: 字符串匹配，就是，n 长主串，m 长匹配串，k 长间隔，问有多少种匹配?

分成 k 组就好，这题可以用来测测你的 KMP 模板哦，数据还可以，就是，就算是朴素匹配也能过。。。

做完我算是真的理解KMP了，对于字符串，有个 \\0 结尾的特性，所以优化的 next 是可行的，但是这种却不行，而且，优化的并不好求匹配数。

 KMP 模板 : http://www.cnblogs.com/haoabcd2010/p/6722073.html

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <vector>
 5 using namespace std;
 6 #define MX 1000005
 7 
 8 int n,m,p;
 9 int ans;
10 int t[MX];
11 vector<int> zu[MX];
12 int net[MX];
13 
14 void Init()
15 {
16     for (int i=0;i<=n;i++)
17         zu[i].clear();
18 }
19 
20 void get_next()
21 {
22     int i=0,j=-1;
23     net[0]=-1;
24     while (i<m)
25     {
26         if (j==-1||t[i]==t[j]) net[++i]=++j;
27         else j = net[j];
28     }
29 
30     for (int i=0;i<=m;i++)
31         printf("%d ",net[i]);
32     printf("\\n");
33 }
34 
35 void KMP(int x)
36 {
37     int i=0,j=0;
38     int len = zu[x].size();
39     while(i<len&&j<m)
40     {
41         if (j==-1||zu[x][i]==t[j])
42         {
43             i++,j++;
44         }
45         else j=net[j];
46         if (j==m)
47         {
48             ans++;
49             j = net[j];
50         }
51     }
52 }
53 
54 int main()
55 {
56     int T;
57     scanf("%d",&T);
58     for(int cas=1;cas<=T;cas++)
59     {
60         scanf("%d%d%d",&n,&m,&p);
61         Init();
62         for (int i=0;i<n;i++)
63         {
64             int x;
65             scanf("%d",&x);
66             zu[i%p].push_back(x);
67         }
68         for (int i=0;i<m;i++)
69             scanf("%d",&t[i]);
70         get_next();
71 
72         ans = 0;
73         for (int i=0;i<p;i++) KMP(i);
74         printf("Case #%d: %d\\n",cas,ans);
75     }
76     return 0;
77 }

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