KMP模板题|hdu 1711

数据大

#include
#include
#define maxn 1000000+5
#define maxn2 10000+5
using namespace std;

int s[maxn];
int p[maxn2];
int Next[maxn2];
void getNext(const int p[],int Next[],int plen) {
    Next[0]=-1;
    int k=-1;
    int j=0;
    while(j=plen) {
        return i-j+1;
    } else {
        return -1;
    }

}

int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        int num1,num2;
        cin>>num1;
        cin>>num2;
        for(int i=0; i

http://acm.split.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30758 Accepted Submission(s): 12930

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