Sequence I

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 211 Accepted Submission(s): 90

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,⋯,bm is 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

Sample Output

Case #1: 2
Case #2: 1

Source

2016中國大學生程序設計競賽（長春）-重現賽

解題思路：
1，字符串匹配，只是文本串每次移動p位而已
2，然後就是要注意，以前kmp做字符轉匹配的時候，模式串的p[m]位置是有'\0'的，然而變成數組的時候
要保證p[m]=0即可，也就是要memset模式串。
3，其實這題數據非常的水，當時暴力n^2的複雜度都能過。（不知道後來數據會不會加強）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000005 ;
int f[maxn] ;
int a[maxn] ;
int b[maxn] ;
int n,m,pp;
int ans ;
void find(int st,int *T,int *P){
    int j = 0 ;
    for(int i=st;i<n;i+=pp){
        while(j&&P[j]!=T[i]){j = f[j];}
        if(P[j]==T[i])j++ ;
        if(j==m){
            ans++ ;
            //j = 0 ;
        }
    }
}
void getfail(int *p){
    f[0] = 0 ;
    f[1] = 0 ;
    for(int i=1;i<m;i++){
        int j = f[i] ;
        while(j&&p[i]!=p[j])j = f[j] ;
        f[i+1] = (p[i]==p[j] ? j+1 : 0) ;
    }
}
int main(){
    int T,cas=1;
    scanf("%d",&T);
    while(T--){
        ans = 0 ;
        scanf("%d%d%d",&n,&m,&pp);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        for(int i=0;i<m;i++)scanf("%d",&b[i]);
        //b[m]=0;
        getfail(b) ;
        for(int i=0;i<pp;i++){
            find(i,a,b);
        }
        printf("Case #%d: %d\n",cas++,ans);
    }
    return 0;
}