題目連結

Problem Description

Mr. Frog has two sequences a1,a2,⋯,an $a_1,a_2,\cdots$

, a n $a_1,a_2,\cdots ,a_n$ and $b_1,b_2,\cdots ,b_m$ and a number p. He wants to know the number of positions q such that sequence $b_1,b_2,\cdots ,b_m$ is exactly the sequence $a_q,a_{q+p},a_{q+2p},\cdots ,a_{q+(m-1)p}$ where $q + (m - 1)p\leq n$ and $q\geq 1$.

Input

The first line contains only one integer $T\leq 100$, which indicates the number of test cases.Each test case contains three lines.The first line contains three space-separated integers $1\leq n\leq 10^6, 1\leq m\leq 10^6$ and $1\leq p\leq 10^6$.The second line contains n integers $a_1,a_2,\cdots ,a_n(1 \leq a_i \leq 10^9)$.the third line contains m integers $b_1,b_2,\cdots ,b_m(1 \leq b_i \leq 10^9)$.

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

AC

• 在a陣列中匹配b，每次間隔p，如果每次間隔1的話，直接一個kmp就可以，但是間隔2以上就要列舉所有的區間進行kmp，要不然會漏掉
• 例如a陣列：1，1，2，2，3，3，b陣列：1, 2, 3 p = 2，如果只是從0開始跑kmp，只會得到1對應a陣列中下標為（0,2,4），但是（1,3,5）就漏掉了
• 所以在滿足可以構成M個數的前提下，保證所有的區間恰好都覆蓋

#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <set>
#include <cstring>
#include <cmath>
#include <algorithm>
#define N 1000005
#define ll  long long
using namespace std;
int a[N], b[N], pre[N];
void get_next(int m) {
    int i = 0, j = -1;
    pre[0] = -1;
    while (i < m) {
        if (b[i] == b[j] || j == -1)
            pre[++i] = ++j;
        else
            j = pre[j];
    }
}

int kmp(int start, int n, int m, int p) {
    int ans = 0;
    get_next(m);
    int i = start, j = 0;
    while (i < n) {
        if (j == -1 || a[i] == b[j])
            j++, i += p;
        else
            j = pre[j];
        if (j == m)
            ans++, j = pre[j];  
    }
    return ans;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int t;
    scanf("%d", &t);
    int Case = 1;
    while (t--) {
        int n, m, p;
        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]);
        }
        int ans = 0;
        // 列舉所有區間 
        for (int i = 0; i < p; ++i) {
            // 保證構成M個數 
            if (i + m * p - p <= n - 1)
                ans += kmp(i, n, m, p);
            else
                break;
        }
        printf("Case #%d: %d\n", Case++, ans);
    } 
    return 0;
}