Number Sequence

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

KMP的模板题

#include <iostream>#include <cstdio>const int MAX=1e6+10;int a[MAX] , b[10010];int next[10010];int kmp(int n ,int m ){   int i = 0 , j =0;   while( i < n)   {      if( j == -1 || a[i] == b[j])       {                    i++;j++;if(j == m)   {           return i-m+1;}        }   else   {        j = next[j];   }   }   return -1;}void getnext( int m){     int i= 0 , j = -1; next[0] = -1 ;  while( i < m) {    if( j == -1 || b[i] == b[j])      {         i++;         j++;         next[i] = j;  }  else  {     j = next[j];  }  } }int main(){       int t;     scanf("%d",&t);     while(t--)    {    int  n , m;    scanf("%d %d",&n,&m);    for(int i = 0; i < n; i++)         scanf("%d",&a[i]);         for(int i = 0; i < m; i++)             scanf("%d",&b[i]);             getnext(m);            printf("%d\n",kmp(n , m));}return 0;}