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
`213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1`

Sample Output
```6-1代码实现：
#include

#include

using namespace std;int next[10005];int a[1000005],b[10005];void getNext(int *b,int m,int n){
int i=0,j=-1;
next[0]=-1;
while(i

{
if(j==-1||b[i]==b[j])
{
i++;j++;next[i]=j;
}
else
{
j=next[j];
}
}}int main(){
int t,m,n;
cin>>t;
while(t--)
{
cin>>m>>n;
for(int i=0;i

{
cin>>a[i];
}
for(int j=0;j

{
cin>>b[j];
}
getNext(b,m,n);
int i=0,j=0,flag=0;
while(i

{
if(j==-1||a[i]==b[j])
{
i++;j++;
}
else
{
j=next[j];
}
if(j==n)
{
flag++;
cout<

<

break;
}
}
if(flag==0)cout<<"-1"<

}
return 0;}

```

#### 最新留言

[***.249.68.14]2022年05月24日 06时48分00秒

## 关于作者

喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！
-- 愿君每日到此一游！