Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 706 Accepted Submission(s): 266

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

【题解】



意思是让你在

a1,a1+p,a1+2p,a1+3p…

a2,a2+p,a2+2p,a2 + 3p..

a3,a3+p,a3+2p,a3 +3p

…

ap,ap+p,ap+2p,ap+3p..

(a右边的东西都是下标;)

这p个序列里面找b数组的匹配数目；

用vector类处理出这个p个数列就好。

剩下的用KMP算法解决。

找完一个匹配之后,j==m。

这个时候让j= f[j];

就能继续找匹配了。

记住就好。不然每次都想好烦。

#include <cstdio>

#include <iostream>

#include <vector>

const int MAXN = 2e6;

const int MAXM = 2e6;

using namespace std;

int p;

vector <int> a[MAXN];

int b[MAXM];

int f[MAXM],ans,n,m;

void input(int &r)

{

    int t = getchar();

    while (!isdigit(t)) t = getchar();

    r = 0;

    while (isdigit(t)) r = r * 10+t-'0', t = getchar();

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    int t;

    input(t);

    for (int q = 1; q <= t; q++)

    {

        for (int i = 0; i <= 1000000; i++)//a[0]也要clear!

            a[i].clear();

        ans = 0;

        input(n); input(m); input(p);

        for (int i = 1; i <= n; i++)

        {

            int x;

            input(x);

            a[(i%p)].push_back(x);

        }

        for (int j = 0; j <= m - 1; j++)

            input(b[j]);

        f[0] = 0; f[1] = 0;

        for (int i = 1; i <= m - 1; i++)//获取失配函数,b数组下表是从0开始的。

        {

            int j = f[i];

            while (j && b[j] != b[i]) j = f[j];

            f[i + 1] = b[j] == b[i] ? j + 1 : 0;

        }

        for (int i = 0; i <= p - 1; i++)//给p个数列找匹配数目

        {

            int j = 0, len = a[i].size();

            for (int k = 0; k <= len - 1; k++)

            {

                while (j && a[i][k] != b[j]) j = f[j];

                if (a[i][k] == b[j])

                    j++;

                if (j == m)

                {

                    ans++;

                    j = f[j];

                }

            }

        }

        printf("Case #%d: %d\n", q, ans);

    }

    return 0;

}