Number Sequence

 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

代码详解：

#include<stdio.h>#include<iostream>using namespace std;int a[1000521],b[10521];int c[10521];//代表next数组//void getext(int *b, int m) {    int i = 0, j = 0;    c[0] = -1; j = c[i];    while(i < m) {        if(j == -1 || b[i] == b[j]) {            c[++i] = ++j;        }        else {            j = c[j];        }    }}//赋值c数组//int Kmp(int* a, int n, int* b, 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 = c[j];        }    }    return -1;}//找到重合的段//int main(){int T;cin >> T ;int N,M;while(T--){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]);}getext(b,M);printf("%d\n",Kmp(a,N,b,M));}return 0;}

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