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 }