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;
}