Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7619    Accepted Submission(s): 3469

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

 

Sample Input

2 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 1 3 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 2 1

 

Sample Output

6 -1

#include
    
     using namespace std;#pragma warning(disable : 4996)const int MAXN = 10005;int Next[MAXN];int text[1000005] = {0};int pat[MAXN] = {0};int n, m;void get_next(){	int i = 0, j = -1;	Next[0] = -1;	while(i < m)	{		if(j == -1 || pat[i] == pat[j])		{			i++;			j++;			Next[i] = j;		}		else		{			j = Next[j];		}	}}int kmp(){	get_next();	int i = 0, j = 0;	while(i < n && j < m)	{		if(j == -1 || text[i] == pat[j])		{			i++;			j++;		}		else		{			j = Next[j];		}	}	if(j >= m)	{		return i - m + 1;	}	else	{		return -1;	}}int main(){	freopen("in.txt", "r", stdin);	int t;	scanf("%d", &t);	while (t--)	{		scanf("%d %d", &n, &m);		for(int i = 0; i < n; i++)		{			scanf("%d", &text[i]);		}		for(int i = 0; i < m; i++)		{			scanf("%d", &pat[i]);		}		printf("%d\n", kmp());	}	return 0;}