HDU 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3492 Accepted Submission(s): 1581

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

HDU 2007-Spring Programming Contest

Recommend

lcy

 

这就是个简单题，直接套模板就可以了。

#include<stdio.h>
#include<string.h>
int ch_a[1000010] , ch_b[10010];
int next[10010] , n , m , cas;  
void getnext( )
{
    memset(next , 0 , sizeof ( next )) ; 
    int i = -1 , j = 0 ;
    next[0] = -1 ; 
    while ( j < m )
    {
        if ( i == -1 || ch_b[i] == ch_b[j] )
        {
            i ++ , j ++ ;
            next[j] = i ;
        }
        else i = next[i] ;
    }
}
int KMP ()
{
    int i = 0 , j = 0  ;
    while ( (i < n ) && (j < m) )
    {
        if ( j == -1 || ch_a[i] == ch_b[j])
        {
            j ++ , i ++ ;
        }
        else j = next[j] ; 
    }
    if ( j == m) return i - m + 1 ; 
    else return -1 ;
}
int main ()
{
    scanf ( "%d" , &cas ) ; 
    while ( cas -- )
    {
        memset( ch_a , 0 , sizeof ( ch_a ) ) ; 
        memset( ch_b , 0 , sizeof ( ch_b ) ) ; 
        scanf ( "%d%d%*c" , &n , &m ) ; 
        for ( int i = 0 ; i < n ; ++ i )
            scanf ( "%d" , ch_a + i ) ;
        for ( int i = 0 ; i < m ; ++ i )
            scanf ( "%d" , ch_b + i ) ; 
        getnext () ; 
        printf ( "%d\n" , KMP() ) ;
    }
    return 0 ; 
}