A - Number Sequence

Given two sequences of numbers : a11, a22, ...... , aNN, and b11, b22, ...... , bMM(1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aKK = b11, aK+1K+1 = b22, ...... , aK+M−1K+M−1 = bMM. If there are more than one K exist, output the smallest one. 

InputThe 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 a11, a22, ...... , aNN. The third line contains M integers which indicate b11, b22, ...... , bMM. All integers are in the range of −1000000,1000000−1000000,1000000. 
OutputFor 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<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
using namespace std;
#define MAXN 1000001
typedef long long LL;
/*
KMP  查找子串首次出现的位置
*/
int s[MAXN],t[MAXN],Next[MAXN];
void kmp_pre(int m)
{
    int j,k;
    j=0;k=-1;Next[0]=-1;
    while(j<m)
    {
        if(k==-1||t[j]==t[k])
            Next[++j] = ++k;
        else
            k = Next[k];
    }
}
int KMP(int n,int m)
{
    int i,j,ans;
    i=j=ans=0;
    kmp_pre(m);
    if(n==1&&m==1)
        return (s[0]==t[0])?1:-1;
    for(i=0;i<n;i++)
    {
        while(j>0&&s[i]!=t[j])
            j = Next[j];
        if(s[i]==t[j])
            j++;
        if(j>=m)
        {
            if(i-m+2>0)
                return i-m+2;
            else
                return -1;
        }
    }
    return -1;
}
int main()
{
    int T,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&s[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&t[i]);
        printf("%d\n",KMP(n,m));
    }
}