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
题意：输入测试次数t，输入主串长度n和模式串长度m,输出模式串在主串中第一次出现的位置，若主串中不含模式串，输出-1。
裸KMP模板题，注意KMP函数最后是匹配完模式串时指针在主串中的位置。还需减掉模式串长度才是第一次出现的位置。
代码如下：
#include<stdio.h>///裸KMP模板
void kmp_pre(int x[],int m,int next[])
{
int i,j;
j=next[0]=-1;
i=0;
while(i<m)
{
while(j!=-1&&x[i]!=x[j])j=next[j];
next[++i]=++j;
}
}
int kmp_count(int x[],int m,int y[],int n)
{
int i=0,j=0,next[10005];
kmp_pre(x,m,next);
while(i<n)
{
while(j!=-1&&x[j]!=y[i])j=next[j];
i++;j++;
if(j>=m)return i+1-m;///注意这里要减去模式串的长度
}
return -1;///如果没有在主串中找到子串，最后输出-1表示找不到
}
int a[1000006],b[10004],n,m,t;
int main()
{
scanf("%d",&t);
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]);
printf("%d
",kmp_count(b,m,a,n));///直接输出第一次查找到完整字符串的位置
}
}