Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25811    Accepted Submission(s): 10908

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
/*标准的 KMP模板模板抄错了，，，结果出不来.qaq **__**用next数组 和using namespace std  会出现编译错误   不解 */#include<stdio.h>#include<algorithm>//using namespace std;int next[10002];int  a[10002],b[1000002];int n,m;void setnext(){int i=0,j=-1;next[0]=-1;while(i<m){if(j==-1||a[i]==a[j]){i++;j++;next[i]=j;}else{j=next[j];}}} int kmp(){setnext();int i=0,j=0;while(i<n){if(j==-1||b[i]==a[j]){i++;j++;}else j=next[j];if(j==m)return i-j+1;}return -1;}int main(){int t,i,j;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=0;i<n;i++)scanf("%d",&b[i]);for(j=0;j<m;j++)scanf("%d",&a[j]);if(kmp()==0)printf("-1\n");elseprintf("%d\n",kmp());}}