hdu 1711 Number Sequence

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

这题是一个最简单的kmp的题目， 子序列的匹配~·

#include<stdio.h>int bbs,n,m;int s[1000010];int a[10010];int next[10010];void tonext()  //得到next数组 {int i=0,j=-1;next[0]=-1;while(i<m){if(j==-1||a[i]==a[j]){i++;j++;if(a[i]==a[j])  //如果还是相等  就继续优化 {next[i]=next[j];}else next[i]=j;}else j=next[j];}}int kmp()  //序列的匹配 { int i=0,j=0;  //2个都设为0 while(i<n&&j<m){if(a[j]==s[i]||j==-1)  //相等 就比较下一个 {i++;j++;}else j=next[j];}if(j>m-1)  //返回第几个开始 return i-m+1;else return -1;}int main(){scanf("%d",&bbs);while(bbs--){scanf("%d%d",&n,&m);if(n<m){printf("-1\n");continue;}for(int i=0;i<n;i++){scanf("%d",&s[i]);}for(int i=0;i<m;i++){scanf("%d",&a[i]);}tonext();int ans=kmp();printf("%d\n",ans);}return 0;}