Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40538 Accepted Submission(s): 16729
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40538 Accepted Submission(s): 16729
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].
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.
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
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
6
-1
C/C++:
1 #include <map> 2 #include <queue> 3 #include <cmath> 4 #include <vector> 5 #include <string> 6 #include <cstdio> 7 #include <cstring> 8 #include <climits> 9 #include <iostream> 10 #include <algorithm> 11 #define INF 0x3f3f3f3f 12 using namespace std; 13 const int my_max = 1e6 + 10; 14 15 int my_len1, my_len2, my_next[my_max], my_num1[my_max], my_num2[my_max]; 16 17 int my_kmp() 18 { 19 for (int i = 1, j = 0; i < my_len2; ++ i) 20 { 21 while (j > 0 && my_num2[i] != my_num2[j]) j = my_next[j]; 22 if (my_num2[i] == my_num2[j]) ++ j; 23 my_next[i + 1] = j; 24 } 25 for (int i = 0, j = 0; i < my_len1; ++ i) 26 { 27 while (j > 0 && my_num1[i] != my_num2[j]) j = my_next[j]; 28 if (my_num1[i] == my_num2[j]) ++ j; 29 if (j == my_len2) return i + 2 - my_len2; 30 } 31 return -1; 32 } 33 34 int main() 35 { 36 int t; 37 scanf("%d", &t); 38 while (t --) 39 { 40 memset(my_num1, 0, sizeof(my_num1)); 41 memset(my_num2, 0, sizeof(my_num2)); 42 memset(my_next, 0, sizeof(my_next)); 43 scanf("%d%d", &my_len1, &my_len2); 44 for (int i = 0; i < my_len1; ++ i) 45 scanf("%d", &my_num1[i]); 46 for (int i = 0; i < my_len2; ++ i) 47 scanf("%d", &my_num2[i]); 48 49 printf("%d ", my_kmp()); 50 } 51 return 0; 52 }