题目：https://vjudge.net/contest/243048#problem
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
直接套模板；
#include <iostream>
#include <stdio.h>
#define maxn 1000005
#define maxm 10010
using namespace std;
int fail[maxm];
int str[maxn];
int temp[maxm];
void getFail(int n) { //n表示短字符串的长度
int i = 0, j = -1; //i表示短字符串当前位置，j表示当前位置之前的部分前缀和后缀相同的个数-1
fail[0] = -1; //fail数组表示当前位置匹配失败后从短字符串之前哪个位置继续匹配
while(i < n) {
if(j==-1 || temp[i]==temp[j]) //temp数组用来存储短字符串
{
++i, ++j;
if(temp[i] != temp[j]) fail[i] = j;
else fail[i] = fail[j];
}
else j = fail[j];
}
}
int kmp(int n, int m) { //n表示长字符串的长度，m表示短字符串的长度 //res表示长字符串中有多少个短字符串
int i = 0, j = 0; //i表示长字符串当前位置，j表示短字符串当前位置
while(i < n) {
if(str[i]==temp[j]) { //temp存储短字符串，str存储长字符串
if(j == m-1) { //当发现长字符串中有一个短字符串时
return i-j+1;
}
++i, ++j;
}
else
{
j = fail[j];
if(j==-1){
i++;j=0;
}
}
}
return -1;
}
int main()
{
int t;
scanf("%d",&t);
int N,M;
while(t--)
{
scanf("%d%d",&N,&M);
for(int i=0;i<N;i++)
scanf("%d",&str[i]);
for(int i=0;i<M;i++)
scanf("%d",&temp[i]);
getFail(M);
int ans= kmp(N,M);
printf("%d\n",ans);
}
return 0;
}