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=0;i<=n;i++)
        zu[i].clear();
}

void get_next()
{
    int i=0,j=-1;
    net[0]=-1;
    while (i<m)
    {
        if (j==-1||t[i]==t[j]) net[++i]=++j;
        else j = net[j];
    }

    for (int i=0;i<=m;i++)
        printf("%d ",net[i]);
    printf("\n");
}

void KMP(int x)
{
    int i=0,j=0;
    int len = zu[x].size();
    while(i<len&&j<m)
    {
        if (j==-1||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=1;cas<=T;cas++)
    {
        scanf("%d%d%d",&n,&m,&p);
        Init();
        for (int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&x);
            zu[i%p].push_back(x);
        }
        for (int i=0;i<m;i++)
            scanf("%d",&t[i]);
        get_next();

        ans = 0;
        for (int i=0;i<p;i++) KMP(i);
        printf("Case #%d: %d\n",cas,ans);
    }
    return 0;
}