Sequence I

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,?,bmis 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

题意：

　　给你a,b两个序列

　　和一个p ，求有多少个 q恰好满足 b1,b2,b3....bm 就是 a[q],a[q+p],a[q+2p]......a[q+(m-1)p];

题解：

　　将a序列，每隔p位置分成一组，这样最多有p组，个数和是n

　　将每组和b序列跑KMP计算答案

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e6+10, M = 1e6, mod = 1e9+7, inf = 2e9;

int T,n,m,p,s[N],t[N],ans = 0;
vector<int > P[N];
int nex[N];
int main()
{
    int cas = 1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&p);
        for(int i = 1; i <= n; ++i) scanf("%d",&t[i]);
        for(int i = 1; i <= m; ++i) scanf("%d",&s[i]);
        for(int i = 0; i < p; ++i) P[i].clear();
        memset(nex,0,sizeof(nex));
        ans = 0;
        int k = 0;
        for(int i=2; i<=m; i++)
        {
            while(k>0&&s[k+1]!=s[i]) k = nex[k];
            if(s[k+1]==s[i])k++;
            nex[i] = k;
        }
        if(p == 1)
        {
            k = 0;
            for(int i=1; i<=n; i++)
            {
                while(k>0&&s[k+1]!=t[i]) k = nex[k];
                if(s[k+1]==t[i]) k++;
                if(k==m) {
                    k = nex[k];
                    ans++;
                }
            }
            printf("Case #%d: ",cas++);
            printf("%d\n",ans);
        }
        else {
            for(int i = 1; i <= n; ++i)
                P[i % p].push_back(t[i]);
            for(int i = 0; i < p; ++i) {
                k = 0;
                for(int j = 0; j < P[i].size(); ++j) {
                    while(k>0&&s[k+1]!=P[i][j]) k = nex[k];
                    if(s[k+1]==P[i][j]) k++;
                    if(k==m) {
                        k = nex[k];
                        ans++;
                    }
                }
            }
            printf("Case #%d: ",cas++);
            printf("%d\n",ans);

        }
    }
}