Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2330    Accepted Submission(s): 874

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

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题意：将bn与an数组每隔p项进行匹配，bn为模式串，an为主串，问有多少种匹配

分析：   每隔p项进行匹配的时候，第一次匹配 0 p 2p 3p 4p

第二次匹配 1 p+1 2p+1 3p+1 4p+1

那么进行p次对主串的扫描完成所有的匹配

代码如下：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

const int N = 1000002;
int Next[N];
int S[N], T[N];
int slen, tlen;//注意每次一定要计算长度
   int n,m,p;
void getNext()
{
    int j, k;
    j = 0; k = -1; Next[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
            Next[++j] = ++k;
        else
            k = Next[k];

}
int KMP_Count()
{
    int ans = 0;
    int i, j = 0;

    if(slen == 1 && tlen == 1)
    {
        if(S[0] == T[0])
            return 1;
        else
            return 0;
    }
    getNext();
  for(int k=0;k<p;k++){
        j=0;
    for(i = k; i < slen; i=i+p)
    {
        while(j>=0 && S[i] != T[j])
            j = Next[j];
        if(j==-1||S[i] == T[j])
            j++;
        if(j == tlen)
        {
            ans++;
            j = Next[j];
        }
    }
  }
    return ans;
}
int main()
{

    int TT,Case=0;
    scanf("%d",&TT);
    while(TT--)
    {
        Case++;
        scanf("%d%d%d",&n,&m,&p);
        for(int i=0;i<n;i++)scanf("%d",&S[i]);
        for(int i=0;i<m;i++)scanf("%d",&T[i]);
        slen=n;
        tlen=m;;
        printf("Case #%d: %d\n",Case,KMP_Count());
    }
    return 0;
}
/*10
3 1 5
1 1 1
1
Case #1: 1
*/