思想：最长公共前后缀
最长公共前后缀

例：abcde
a       0
ab      0
aba     1 //(a)
abab    2 //(ab)
ababc   0

得到最大公共前后缀

void getnext() 
{
    int i=1;
    int j=0;
    next[0]=0;
    while(i<m)
    {
        if(t[j]==t[i])
        {
            next[i++]=++j;
        }
        else if(!j)
        {
            i++;
        }
        else
        {
            j=next[j-1];
        }
    }
}

例题：
A – 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

#include<iostream>
#include<algorithm>
#include<stdio.h>
//using namespace std;
const int maxn = 1e6+20;
int next[maxn];
int n,m;
char s[maxn];//母串 
char t[maxn];//子串
void getnext() 
{
    int i=1;
    int j=0;
    next[0]=0;
    while(i<m)
    {
        if(t[j]==t[i])
        {
            next[i++]=++j;
        }
        else if(!j)
        {
            i++;
        }
        else
        {
            j=next[j-1];
        }
    }
} 
int kmp()
{
    int i=0;
    int j=0;
    while(i<n&&j<m)
    {
        if(s[i]==t[j])
        {
            i++;
            j++;
        }
        else if(!j)
        {
            i++;
        }
        else
        {
            j=next[j-1];
        }
    }
    if(j==m) return i-m+1;
    else return -1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s %s",s,t);
        n = strlen(s),m = strlen(t);
        getnext();
        printf("%d\n",kmp());
    }
    return 0;
}