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

 

裸KMP算法，也是我第一道KMP算法题

 

#include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              using namespace std;#define inf 0xffffff#define maxn 1000005#define maxm 10005int M,N;int S[maxn],T[maxm];void get_next(int T[],int *next){ next[0] = -1; int i = 0,j = -1; while(i < M){ if(j == -1 || T[i] == T[j]){ ++i,++j; if(T[i] != T[j]) next[i] = j; else next[i] = next[j]; } else j = next[j]; }}int KMP(int S[],int T[]){ int i = 0; int j = 0; int next[maxm]; get_next(T,next); while(i < N && j < M) { if(j == -1 || S[i] == T[j]) { j++;i++; } else j = next[j]; } if( j == M ) return i-M+1; else return -1;}int main(){ int t; scanf("%d",&t); while(t--) { cin>>N>>M; for(int i = 0; i < N; i++) scanf("%d",&S[i]); for(int i = 0; i < M; i++) scanf("%d",&T[i]); cout<
              
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我也不想说什么了1个月没有刷题出现缺少分号，主函数打错和在主函数中定义大数组导致程序崩溃我也是醉了，暑假加油！！！！！