HDU 5918 Sequence I kmp算法+虽然暴力也能过

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

26 3 11 2 3 1 2 31 2 36 3 21 3 2 2 3 11 2 3

 

Sample Output

Case #1: 2Case #2: 1

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛

解题思路：
1，字符串匹配，只是文本串每次移动p位而已
2，然后就是要注意，以前kmp做字符转匹配的时候，模式串的p[m]位置是有'\0'的，然而变成数组的时候
要保证p[m]=0即可，也就是要memset模式串。
3，其实这题数据非常的水，当时暴力n^2的复杂度都能过。（不知道后来数据会不会加强）

#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;}