hdu 1711 Number Sequence

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

题意概括：找出第二个串在第一个串中第一次出现的位置，如果没有输出-1。

解题思路：KMP跑一遍，如果最后j等于第二个串的长度证明出现过输出i-j+1就是第一次出现的位置（找到第二个串完全出现的时候循环结束），如果不等于输出-1.

  代码：

#include<stdio.h>#include<string.h>#define N 1001000#define M 10100int a[N];int b[M];int next[M];int m,n;void GetNext(){int i=0;int j=-1;next[0]=-1;while(i<m){if(j==-1||b[i]==b[j]){i++;j++;next[i]=j;}else{j=next[j];}}}void KMP(){GetNext();int i=0;int j=0;while(i<n&&j<m){if(j==-1||a[i]==b[j]){i++;j++;}else{j=next[j];}}if(j==m)printf("%d\n",i-j+1);elseprintf("-1\n");}int main(){int t,i,j;scanf("%d",&t);while(t--){memset(a,0,sizeof(a));memset(b,0,sizeof(b));scanf("%d%d",&n,&m);for(i=0;i<n;i++){scanf("%d",&a[i]);}for(i=0;i<m;i++){scanf("%d",&b[i]);}KMP();}return 0;}