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
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

题解：kmp算法，模板题；

全当做记模板。

#include <stdio.h>#include <stdlib.h>#include <string.h>int  s[1000010], t[10010];int next[10010]; int Kmp(int * s, int n, int * t, int m) {    int i = 0, j = 0;    while(i < n)     {        if(j == -1 || s[i] == t[j])         {            ++i; ++j;            if(j == m)             {                return i - m + 1;             }        }        else         {            j = next[j];         }    }    return -1;}void getnext(int *t, int m) {    int i = 0, j = 0;    next[0] = -1; j = next[i];    while(i < m) {        if(j == -1 || t[i] == t[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",&s[i]);        for(int i=0;i<m;i++)        scanf("%d",&t[i]);        getnext(t, m);        printf("%d\n", Kmp(s, n, t, m));    }    return 0;}