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

题目大意：

给定两个数组，问能不能再第一个数组中匹配得到第二个数组，如果可以，那么输出最早匹配的起始位置，否则输出-1。

C++

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10010;
int arr1[N*100],arr2[N],nt[N];
int n,m,s;
void getnt()
{
    int i = 0, j = -1;
    nt[0] = -1;
    while(i < m)
    {
        if(j == -1 || arr2[i] == arr2[j])
        {
            i++, j++;
            nt[i] = j;
        }
        else j = nt[j];
    }
}
int kmp()
{
    getnt();
    int i = 0, j = 0;
    s=-1;
    while(i < n)
    {
        if(j == -1 || arr1[i] == arr2[j])
            i++, j++;
        else
            j = nt[j];
        if(j == m)
        {
            s=i-m+1;
            break;
        }
    }
    return s;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++)
            scanf("%d", &arr1[i]);
        for(int i = 0; i < m; i++)
            scanf("%d", &arr2[i]);
         if(n < m) printf("-1\n");
        else cout<<kmp()<<endl;
    }
    return 0;
}