HDU

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

题解：

直接把多个串拿出来单独跑KMP。

代码：

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 1e6+10;

int A[MAXN];
int B[MAXN];
int C[MAXN];
int next1[MAXN];

void getNext(int M){

    int k = -1;
    int j = 0;
    next1[0] = -1;//next数组最好是从0开始要不然会很麻烦 
    while(j<M){
        if(k == -1 || B[j] == B[k]){
            if(B[j+1] == B[k+1]){ 
                next1[++j] = next1[++k];
            }
            else next1[++j] = ++k;
        }
        else {
            k = next1[k];
        }
    }
}

int Find(int N,int M,int st){//返回匹配到的位置 
    int temp1 = st,temp2 = 0;
    while(temp1<N && temp2<M){
        if(temp2 == -1 || C[temp1] == B[temp2]){
            temp1++;temp2++;
        }
        else{
            temp2 = next1[temp2];
        }
    } 
    if(temp2 == M)return temp1-temp2;
    else return -1;  
}


int main(){
	
	int T,N,M,P;
	scanf("%d",&T);
	for(int _=1 ; _<=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]);
		int num,sum = 0;
		getNext(M);
		for(int i=0 ; i<P ; ++i){
			num = 0;
			for(int j = i;j < N && i+(M-1)*P < N;j += P)C[num++] = A[j];
           	int t = Find(num,M,0);
           	while(t != -1){
           		++sum;
				t = Find(num,M,t+1);	
		   	}
		}
		printf("Case #%d: %d
",_,sum);
	}
	
	return 0;
}