題目：

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. 

輸入：

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]. 

輸出：

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

樣例輸入：

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

樣例輸出：

6
-1

輸入兩個陣列，從第一個數組裡找到一個K位置，使得滿足題目中的要求。

AC程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=1000005;
const int maxx=10005;
int a[maxn],b[maxx],net[maxx];//不能用next作為陣列名，next是個關鍵字
int n,m;
void getnet()
{
    net[0]=-1;
    int k=-1;
    int j=0;
    while(j<m)
    {
        if(k==-1||b[j]==b[k])
        {
            ++j;
            ++k;
            if(b[j]!=b[k])//陣列優化
                net[j]=k;
            else
                net[j]=net[k];
        }
        else
        {
            k=net[k];
        }
    }
}
int Kmpserch()
{
    getnet();
    int i=0;
    int j=0;
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
        {
            j=net[j];
        }
    }
    if(j==m)
        return i-j+1;//之所以要加1的原因是第一個a[K]要等於b[1],如果等於b[0],就不用加1了
    else
        return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        if(Kmpserch()==-1)
            printf("-1\n");
        else
            printf("%d\n",Kmpserch());
    }
    return 0;
}