标签：
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
Source
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1358 3336 3746 1251 2222 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 }
杭电1711--Number Sequence(Kmp → → 利用Next数组求串在串中的位置)
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原文地址：http://www.cnblogs.com/fengshun/p/4713770.html
