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

Problem Description Mr. Frog has two sequences $a_1,a_2,⋯,a_n$ and $b_1,b_2,⋯,b_m$ and a number $p$. He wants to know the number of positions q such that sequence $b_1,b_2,⋯,b_m$

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 $1≤n≤10^6,1≤m≤10^6$ and $1≤p≤10^6$.

The second line contains n integers $a_1,a_2,⋯,a_n(1≤a_i≤10^9)$.

the third line contains m integers $b_1,b_2,⋯,b_m(1≤b_i≤10^9)$

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

題意： 給$n,m,p$，再給兩個數字串$a,b$，$a$的長度為$n$，$b$的長度為$m$。問有多少個$q>=1$滿足 $b_1,b_2,⋯,b_m$ 與 $a_q,a_{q+p},a_{q+2p},⋯,a_{q+(m−1)p}$ 完全一致 題解： kmp，把$a$中的元素按照$i$歸類，然後對於每個類當成一個串來用kmp統計這個串中有多少個與$b$完全一致的子串。 注意：用kmp統計符合條件的子串個數的時候，每匹配到一個子串就要將指標向下移動一格。不然會重複計算。

#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];///記住要後移
        }
    }
    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: %d\n",CASE,ans);
    return 0;
}
int LiangJiaJun(){
    int T;scanf("%d",&T);
    for(int i=1;i<=T;i++)w33ha(i);
    return 0;
}