Hdoj 1711 KMP Number Sequence

Problem descriptiongiven 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.

Inputthe 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 [-1000000,100 0000].

Outputfor each test case, You shoshould 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
#include <stdio.h>int n, m;int a[1000005];int b[10005];int f[10005];void getfail(){    f[0] = 0;    f[1] = 0;    for (int i = 1; i < m; i++)    {        int j = f[i];        while (j && b[i] != b[j])            j = f[j];        f[i+1] = b[j] == b[i] ? j + 1 : 0;    }}int main(){    int flag, i, j, t;    scanf("%d",&t);    while (t--)    {        scanf("%d %d",&n,&m);        for (i = 0; i < n; i++)            scanf("%d",&a[i]);        for (i = 0; i < m; i++)            scanf("%d",&b[i]);        getfail();        for (i = 0;i <= m; i++) printf("%d ",f[i]);  puts("");        flag = 1;        j = 0;        for (i = 0;i < n;i++)        {            while (j && b[j] != a[i])                j = f[j];            if (b[j] == a[i])                j++;            if (j == m)            {                flag = 0;                printf("%d\n",i - m + 1);            }        }        if (flag)            puts("-1");    }    return 0;}