HDU——T 1711 Number Sequence

http://acm.hdu.edu.cn/showproblem.php?pid=1711

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29129    Accepted Submission(s): 12254

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

 

模板练习题

 #include <algorithm>
 #include <cstdio>

 using namespace std;

 const int N(+);
 const int M(+);
 int l1,l2,a[N],b[N],p[N];

 inline void Get_next()
 {
     for(int j=,i=;i<=l2;i++)
     {
         if(j>&&b[i]!=b[j+]) j=p[j];
         if(b[i]==b[j+]) j++;
         p[i]=j;
     }
 }
 inline void kmp()
 {
     for(int j=,i=;i<=l1;i++)
     {
         if(j>&&a[i]!=b[j+]) j=p[j];
         if(a[i]==b[j+]) j++;
         if(j==l2)
         {
             printf("%d\n",i-j+);
             return ;
         }
     }
     puts("-1");
 }

 int main()
 {
     int t;scanf("%d",&t);
     for(;t--;)
     {
         scanf("%d%d",&l1,&l2);
         for(int i=;i<=l1;i++) scanf("%d",a+i);
         for(int i=;i<=l2;i++) scanf("%d",b+i);
         Get_next();
         kmp();
     }
     return ;
 }