Sequence I

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

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <vector>
 using namespace std;
 #define MX 1000005
 
 int n,m,p;
 int ans;
 int t[MX];
 vector<int> zu[MX];
 int net[MX];
 
 void Init()
 {
     for (int i=;i<=n;i++)
         zu[i].clear();
 }
 
 void get_next()
 {
     int i=,j=-;
     net[]=-;
     while (i<m)
     {
         if (j==-||t[i]==t[j]) net[++i]=++j;
         else j = net[j];
     }
 
     for (int i=;i<=m;i++)
         printf("%d ",net[i]);
     printf("\n");
 }
 
 void KMP(int x)
 {
     int i=,j=;
     int len = zu[x].size();
     while(i<len&&j<m)
     {
         if (j==-||zu[x][i]==t[j])
         {
             i++,j++;
         }
         else j=net[j];
         if (j==m)
         {
             ans++;
             j = net[j];
         }
     }
 }
 
 int main()
 {
     int T;
     scanf("%d",&T);
     for(int cas=;cas<=T;cas++)
     {
         scanf("%d%d%d",&n,&m,&p);
         Init();
         for (int i=;i<n;i++)
         {
             int x;
             scanf("%d",&x);
             zu[i%p].push_back(x);
         }
         for (int i=;i<m;i++)
             scanf("%d",&t[i]);
         get_next();
 
         ans = ;
         for (int i=;i<p;i++) KMP(i);
         printf("Case #%d: %d\n",cas,ans);
     }
     return ;
 }