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

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<cstdio>
 5 using namespace std;
 6 
 7 int nxt[1000005],s[1000005],p[10005]; 
 8 int T,n,m;
 9 
10 void getNext(){
11     nxt[0]=-1;
12     for(int i=0;i<m;i++){
13         int k=nxt[i];
14         while(k!=-1&&p[i]!=p[k]) k=nxt[k];
15         nxt[i+1]=k+1;
16     }
17 }
18 
19 int kmp(){
20     getNext();
21     int i=0,j=0;
22     while(i<n&&j<m){
23         while(j!=-1&&s[i]!=p[j]) j=nxt[j];
24         ++i;
25         ++j;   
26     }
27     if(j==m) return i-j+1;
28     return -1;
29 }
30 
31 int main()
32 {   scanf("%d",&T);
33     while(T--){
34         scanf("%d%d",&n,&m);
35         for(int i=0;i<n;i++) scanf("%d",&s[i]);
36         for(int i=0;i<m;i++) scanf("%d",&p[i]);
37         int ans=kmp();
38         printf("%d\n",ans);
39     }
40     return 0;
41 }

Number Sequence HDU - 1711