HDU 1711 Number Sequence

题目：

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
解题思路：
KMP模板题
代码实现：   

#include<stdio.h>
int a[1000010],b[10010];
int la,lb;
void get_next(int b[],int next[]);
int kmp(int a[],int b[]);
int main()
{
    int i,t,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&la,&lb);
        for(i=0;i<la;i++)
            scanf("%d",&a[i]);
        for(i=0;i<lb;i++)
            scanf("%d",&b[i]);
        k=kmp(a,b);
        if(k>=0)
            printf("%d
",k+1);
        else//不匹配返回-1 
            printf("-1
");
    }
    return 0;
}
int kmp(int a[],int b[])
{
    int i,j;
    int next[10010];
    
    get_next(b,next);
    i=0;
    j=0;
    while(i < la && j < lb)
    {
        if(j==-1 || a[i] == b[j])
        {
            i++;
            j++;
        }
        else
            j=next[j];
    }
    //printf("%d
",i-lb);
    if(j >= lb)
        return i-lb;//返回i-匹配串的长度 
    else
        return -1;
} 
void get_next(int b[],int next[])
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;
    while(i < lb)
    {
        if(j==-1 || b[i] == b[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
    /*for(i=0;i<lb;i++)
        printf("%d ",next[i]);
    printf("
");*/
}