HDOJ 1711Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15622    Accepted Submission(s): 6877

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

 

#include <cstdio>const int N = 1000010;const int M = 10010;int ary[N];int p[M], next[M];int n, m;void get_next() {int i = 0, j = -1;next[0] = -1;while (i < m) {if (j == -1 || p[i] == p[j]) {++i, ++j;if (p[i] != p[j])next[i] = j;elsenext[i] = next[j];}elsej = next[j];}}int KMP() {int i = 0, j = 0;while (i < n && j < m) {if (ary[i] == p[j])++i, ++j;else {if (next[j] == -1)++i, j = 0;elsej = next[j];}}if (j == m)return i - j + 1;elsereturn -1;}int main() {int T;scanf("%d", &T);while (T--) {scanf("%d%d", &n, &m);for (int i = 0; i < n; ++i)scanf("%d", ary + i);for (int i = 0; i < m; ++i)scanf("%d", p + i);get_next();printf("%d\n", KMP());}return 0;}