### 【HDU 1711】Number Sequence 【KMP 模板】

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

``````#include<cstring>
#include<cstdio>
using namespace std ;
typedef long long LL ;

const int MAXN = 1000000+10 ;
const int MAXM = 1e5 ;
const int mod  = 1e9+7;

int s[1000000+10],t[10000+10];
int n,m;
int nexts[10000+10];
void get(){
int i,j;i=j=0;
nexts[0]=-1;j=-1;
while(i<m){
if(j==-1||t[i]==t[j])  nexts[++i]=++j;
else j=nexts[j];
}
}
int Kmp(){
int i,j;i=j=0;
while(i<n){
if(j==-1||s[i]==t[j]) {
++i;++j;
if(j==m) return i-m+1;
}else j=nexts[j];
}
return -1;
}
int main(){
int z; scanf("%d",&z);
while(z--){
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]);
get();
printf("%d\n",Kmp());
}
return  0;
}``````