Number Sequence

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21383 Accepted Submission(s): 9163

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
求连续字串的开始点位置
kmp模板套用

#include<cstdio>#include <iostream>#include<cstring>using namespace std;int a[1000010] , b[10010];int nextt[10010] , n , m ;void get_next( ){    memset(nextt , 0 , sizeof ( nextt )) ;    int i = -1 , j = 0 ;    nextt[0] = -1 ;    while ( j < m )    {        if ( i == -1 || b[i] == b[j] )        {            i ++ , j ++ ;            nextt[j] = i ;        }        else i = nextt[i] ;    }}int kmp(){    int i = 0 , j = 0  ;    while ( (i < n ) && (j < m) )    {        if ( j == -1 || a[i] == b[j])        {            j ++ , i ++ ;        }        else j = nextt[j] ;    }    if ( j == m) return i - m + 1 ;    else return -1 ;}int main (){    int c;    scanf ( "%d" , &c ) ;    while ( c -- )    {        memset(a , 0 , sizeof ( a ) ) ;        memset(b , 0 , sizeof ( b ) ) ;        scanf ( "%d%d" , &n , &m ) ;        for ( int i = 0 ; i < n ; ++ i )            scanf ( "%d" , a + i ) ;        for ( int i = 0 ; i < m ; ++ i )            scanf ( "%d" , b + i ) ;        get_next () ;        printf ( "%d\n" , kmp() ) ;    }    return 0 ;}