HDU1711-Number Sequence

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

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

#include<cstdio>///HDU1711-Number Sequenceusing namespace std;int a[1000005],b[10005],next[10005];int n,m;int getnext(){    int j=1,k=0;    next[1]=0;    while(j<=m)    {        if(k==0||b[j]==b[k])        {            j++;k++;            next[j]=k;        }        else            k=next[k];    }}int kmp(){    getnext();    int i=1,j=1;    while(i<=n)    {        if(j==0||a[i]==b[j])        {            i++;            j++;        }        else            j=next[j];        if(j==m+1)        {            return i-m;        }    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&n,&m);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        for(int j=1;j<=m;j++)        {            scanf("%d",&b[j]);        }        printf("%d\n",kmp());    }    return 0;}