HDU--1711--Number Sequence

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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define N 110000
#define M 11000
int a[N], b[M], t[M];
using namespace std;
void getnext(int m)
{
    t[0]=-1;
    int i=0, j=-1;
    while(i<m)
    {
        if(j==-1 || b[i]==b[j])
        {
            i++;
            j++;
            if(b[i]!=b[j])
                t[i]=j;
            else
                t[i]=t[j];
        }
        else
            j=t[j];
    }
}
int kmp(int n, int m)
{
    int j=0, i=0;
    while(i<n && j<m)
    {
        if(a[i]==b[j] || j==-1)
        {
            i++;
            j++;

        }
        else
            j=t[j];

    }
    if(j==m)
        return i-m+1;
    else
        return -1;
}
int main()
{
    int i, n, T, m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=0; i<m; i++)
        {
            scanf("%d",&b[i]);
        }
        if(m>n)
            printf("-1\n");
        else
        {
            getnext(m);
            printf("%d\n", kmp(n,m));
        }

    }
    return 0;
}