HDU 1711 Number Sequence //简单kmp

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10570    Accepted Submission(s): 4813

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest

 

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#include <stdio.h>int a[10005], b[1000005], next[10005];void findnext(int m){    int i, j;    i=0;    j=-1;    next[0] = -1;    while(i<m)    {        if(j==-1 || a[i]==a[j])        {            i++;            j++;            next[i] = j;        }        else                    j = next[j];            }}int main(){    int t;    int n, m;    int i, j, flag;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        for(i=0; i<n; i++)            scanf("%d", &b[i]);        for(i=0; i<m; i++)            scanf("%d", &a[i]);        findnext(m);        flag = 0;        i = 0;        j = 0;        while(i<n)        {            if(j==-1 || b[i]==a[j])            {                i++;                j++;                if(j>=m)                {                    printf("%d\n", i-j+1);                                        flag = 1;                    break;                }            }            else                j = next[j];        }        if(flag == 0)            printf("-1\n");    }    return 0;}