HDU5918(KMP)

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 729    Accepted Submission(s): 277

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

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

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛

 

 题意：问从a串中每隔p取字，有多少个可与b串匹配

思路：用kmp可计算连续的a串中有多少个b串，可将每p个数字取出组成连续的新串，共可取出p条新串，对每条新串kmp再求和即可

//2016.10.08
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int N = 1000005;
const int inf = 0x3f3f3f3f;
int a[N], b[N], nex[N],n, m, p;

void pre_kmp(int b[], int m)//得到next数组
{
    int i, j;
    j = nex[0] = -1;
    i = 0;
    while(i < m)
    {
        while(j != -1 && b[i]!= b[j])j = nex[j];
        nex[++i] = ++j;
    }
}

int kmp(int a[], int n, int b[], int m)
{
    int ans = 0;
    pre_kmp(b, m);
    for(int pos = 0; pos < p; pos++)
    {
        int i = pos, j = 0;
        while(i < n)
        {
            while(j != -1 && a[i] != b[j])j = nex[j];
            i += p; j++;
            if(j >= m){ans++; j = nex[j];}
        }
    }
    return ans;
}

int main()
{
    int T, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &m, &p);
        for(int i = 0; i < n; i++)scanf("%d", &a[i]);
        for(int i = 0; i < m; i++)scanf("%d", &b[i]);
        printf("Case #%d: %d
", ++kase, kmp(a,n,b,m));
    }

    return 0;
}