HDOJ 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19127    Accepted Submission(s): 8218

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

 

此题可以作为KMP算法的入门题目，题意为输入一个母串和一个模板串，在母串中找到一个连续的和模板串一样的子串，再输出其开头索引，若没有就输出-1。

#include <cstdio>#include <cstring>#include <iostream>int a[1000000+10],b[10000+10];int nex[10000+10];int n,m;void getNext(){int j,k;j = 0,k = -1;nex[0] = -1;while(j < m){if(k == -1 || b[j] == b[k])nex[++j] = ++k;else k = nex[k];}}int KMP_Index(){int i=0, j=0;getNext();while(i<n && j<m){if(j == -1 || a[i] == b[j])i++,j++;else j = nex[j];}if(j == m) return i-m+1;else return -1;}int main(){int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(int i=0; i<n; i++)scanf("%d",&a[i]);for(int i=0; i<m; i++)scanf("%d",&b[i]);printf("%d\n",KMP_Index());}return 0;}