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.
InputThe 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].
OutputFor 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
將字符換成數字的KMP

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 int s[10005],ss[1000005],Next[10005];
 8 int T,n,m;
 9 
10 void getnext(int* s)
 
11 {
12     Next[0]=0;
13     Next[1]=0;
14     for(int i=1;i<m;i++)
15     {
16         int j=Next[i];
17         while(j&&s[i]!=s[j])
18             j=Next[j];
19         if(s[i]==s[j])
20             Next[i+1]=j+1;
21             else
22                 Next[i+1]=0;
23     }
24 }
25 
26 int
 
 main()
27 {
28     scanf("%d",&T);
29     while(T--)
30     {
31         int flag=0;
32         memset(Next,0,sizeof(Next));
33         memset(ss,0,sizeof(ss));
34         memset(s,0,sizeof(s));
35         scanf("%d%d",&n,&m);
36         for(int i=0;i<n;i++)
37             scanf("%d",&ss[i]);
38         for(int i=0;i<m;i++)
39             scanf("%d",&s[i]); 
40         getnext(s);
41         int j=0;
42         for(int i=0;i<n;i++)
43         {
44             while(j&&s[j]!=ss[i])
45                 j=Next[j];
46             if(ss[i]==s[j])
47                 j++;
48             if(j==m)
49             {
50                 flag=1;
51                 printf("%d\n",i-m+2);
52             }
53             
54         }
55         if(flag==0)
56             printf("-1\n");
57     }
58     
59     
60     return 0;
61 }

HDU - 1711 Number Sequence