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
//low方法应该是一点一点的找，但是，运用KMP算法会变牛逼。。（KMP竟然还可以用到数组中！！(⊙o⊙)
//这题很有趣，让我知道算法更重要的是思想，不要有局限自己思维
#include <iostream>
#include <cstdio> using namespace std;
int data1[],data2[];
int nex[],n,m; void getnext()
{
int i=,j=-;
nex[]=-;
while(i<m)
{
if(j==-||data2[i]==data2[j])
{
i++;j++;
if(data2[i]==data2[j])
nex[i]=nex[j];
else
nex[i]=j;
}
else
j=nex[j];
}
} int kmp()
{
int i=,j=;
getnext();
while(i<n)
{
if(j==-||data1[i]==data2[j])
{
i++;j++;
}
else
{
j=nex[j];// j=nex[i]这里又记错了。。。。
}
if(j==m)
return i-m+;
}
return -;
} int main()
{
int l;
cin>>l;
while(l--)
{
cin>>n>>m;
for(int i=;i<n;i++)
scanf("%d",&data1[i]);
for(int i=;i<m;i++)
scanf("%d",&data2[i]);
getnext();
int ans=kmp();
cout<<ans<<endl; }
return ;
}