Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15068    Accepted Submission(s):
6606

Problem Description

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

Source

HDU
2007-Spring Programming Contest

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RE:从s串中如果能找出p串，则输出p串在s串的位置;

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 int a[1000005], b[10005];
 6 int p[1000005];
 7 int n, m;
 8 void Getp()                 //对  b[]  求   p[] ;
 9 {
10     int i = 0, j = -1;
11     p[i] = j;
12     while(i < m)
13     {
14         if(j == -1 || b[i] == b[j])
15         {
16             i++; j++;
17             p[i] = j;
18         }
19         else
20             j = p[j];
21     }
22 }
23 int Kmp()
24 {
25     int i = 0, j = 0;
26     while(i < n)
27     {
28         if(j == -1 || a[i] == b[j])
29         {
30             i++, j++;
31             if(j == m)
32             return i -j + 1;
33         }
34         else
35             j = p[j];
36     }
37     return -1;
38 }
39 int main()
40 {
41     int t;
42     scanf("%d", &t);
43     while(t--)
44     {
45         int i;
46         scanf("%d %d", &n, &m);
47         for(i = 0; i < n; i++)
48             scanf("%d", &a[i]);
49         for(i = 0; i < m; i++)
50             scanf("%d", &b[i]);
51         Getp();
52         printf("%d\n", Kmp());
53     }
54     return 0;
55 }