[HDU5918]Sequence I

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

Problem Description
Mr. Frog has two sequences $a_1,a_2,\cdots ,a_n$ and $b_1,b_2,\cdots ,b_m$ and a number $p$. He wants to know the number of positions q such that sequence $b_1,b_2,\cdots ,b_m$ is exactly the sequence $a_q,a_{q+p},a_{q+2p},\cdots ,a_{q+(m-1)p}$ where $q+(m-1)p\le n$ and $q\ge 1$.

Input
The first line contains only one integer $T\le 100$, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers $1\le n\le 10^6,1\le m\le 10^6$ and $1\le p\le 10^6$.

The second line contains n integers $a_1,a_2,\cdots ,a_n(1\le a_i\le 10^9)$.

the third line contains m integers $b_1,b_2,\cdots ,b_m(1\le b_i\le 10^9)$.

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：
给$n,m,p$，再给两个数字串$a,b$，$a$的长度为$n$，$b$的长度为$m$。问有多少个$q>=1$满足 $b_1,b_2,\cdots ,b_m$ 与 $a_q,a_{q+p},a_{q+2p},\cdots ,a_{q+(m-1)p}$ 完全一致
题解：
kmp，把$a$中的元素按照$i$归类，然后对于每个类当成一个串来用kmp统计这个串中有多少个与$b$完全一致的子串。
注意：用kmp统计符合条件的子串个数的时候，每匹配到一个子串就要将指针向下移动一格。不然会重复计算。

#include<bits/stdc++.h>
#define LiangJiaJun main
#define MOD 19991227
using namespace std;
int n,m,p;
int a[1000004],b[1000004],nt[1000004];
vector<int>ov[1000004];
int getnext(){
    memset(nt,0,sizeof(nt));
    int j=0;
    for(int i=2;i<=m;i++){
        while(j>0&&b[j+1]!=b[i])j=nt[j];
        if(b[j+1]==b[i])j++;
        nt[i]=j;
    }
    return 0;
}
int calc(int st){
    int j=0,res=0;
    for(int i=0;i<ov[st].size();i++){
        while(j>0&&b[j+1]!=ov[st][i])j=nt[j];
        if(b[j+1]==ov[st][i])j++;
        if(j==m){
            res++;
            j=nt[j];///记住要后移
        }
    }
    return res;
}
int w33ha(int CASE){
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=m;i++)scanf("%d",&b[i]);
    getnext();
    for(int i=0;i<p;i++)if(ov[i].size()>0)ov[i].clear();
    for(int i=1;i<=n;i++)ov[i%p].push_back(a[i]);
    int ans=0;
    for(int i=0;i<p;i++){
        if(ov[i].size()>=m)ans+=calc(i);
    }
    printf("Case #%d: %dn",CASE,ans);
    return 0;
}
int LiangJiaJun(){
    int T;scanf("%d",&T);
    for(int i=1;i<=T;i++)w33ha(i);
    return 0;
}

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