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Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29260 Accepted Submission(s): 12300
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
#include <stdio.h> #include <string.h> int S[1000005],T[1000005]; //S是目标串，T是字串 int next[2000000],l1,l2; void Getnext() {int i=0,j=-1;next[0]=-1; j=next[i];while(i<l2){if(j==-1||T[i]==T[j]){next[i+1]=j+1;i++; j++;}else{j=next[j];}} } int Kmp() {Getnext();int i=0,j=0;while(i<l1){if(j==-1 || S[i]==T[j]){i++; j++; if(j == l2)return i-l2+1;}else{j=next[j];}} return -1; } int main() {int x;scanf("%d",&x);while( x--){ scanf("%d%d",&l1,&l2); for(int i=0; i<l1; i++)scanf("%d",&S[i]);for(int i=0; i<l2; i++)scanf("%d",&T[i]);printf("%d\n", Kmp());} return 0; }