Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 27028    Accepted Submission(s): 11408

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

题目链接：

分析：KMP裸题，自己看吧，不会的看我博客详解！此题有道坑点就是读入不能用cin读入，很容易T！

纯粹要看运气才会过QAQ

优化以后：

 

 速度快了将近3.5s，scanf大法好啊

下面给出AC代码：

 

1 #include 
      
         2 using namespace std;  3 const int N=1000050;  4 inline int read()  5 {  6     int x=0,f=1;  7     char ch=getchar();  8     while(ch<'0'||ch>'9')  9     { 10         if(ch=='-') 11             f=-1; 12         ch=getchar(); 13     } 14     while(ch>='0'&&ch<='9') 15     { 16         x=x*10+ch-'0'; 17         ch=getchar(); 18     } 19     return x*f; 20 } 21 int kmpnext[N]; 22 int s[N],t[N];///s为主串，t为模式串 23 int slen,tlen;///slen为主串的长度，tlen为模式串的长度 24 inline void getnext() 25 { 26     int i,j; 27     j=kmpnext[0]=-1; 28     i=0; 29     while(i
       
        0&&s[i]!=t[j]) 82             j=kmpnext[j]; 83         if(s[i]==t[j]) 84             j++; 85         if(j==tlen) 86         { 87             ans++; 88             j=kmpnext[j]; 89         } 90     } 91     return ans; 92 } 93 int T; 94 int main() 95 { 96     T=read(); 97     while(T--) 98     { 99         slen=read();100         tlen=read();101         for(int i=0;i