hdu——1711——Number Sequence

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

 

#include <iostream>using namespace std;int next[100000];int n,m;int s1[1000000];int s2[10000];void getnext(){    int j=0;    int k=-1;    next[0]=-1;    while(j<m-1)    {        if(k==-1||s2[j]==s2[k])        {           j++;           k++;           next[j]=k;        }        else            k=next[k];    }}int kmp(){    int i=0;    int j=0;    while(i<n)    {        if(j==-1||s1[i]==s2[j])        {            i++;            j++;        }        else        {            j=next[j];        }        if(j==m)        {            return i-m+1;        }    }    return -1;}int main(){    int k;    cin>>k;    while(k--)    {        cin>>n>>m;        for(int i=0;i<n;i++)        {            //cin>>s1[i];            scanf("%d",&s1[i]);        }        for(int i=0;i<m;i++)        {            //cin>>s2[i];            scanf("%d",&s2[i]);        }        getnext();        if(n<m)            cout<<"-1"<<endl;        else            cout<<kmp()<<endl;    }    return 0;}