Hdu1711:number Sequence

Problem Descriptiongiven-sequences of numbers:a[1], a[2], ..., a[n], and b[1], b[2], ..., b[m] (1 <= M &L t;= 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 is more than one K exist, output the smallest one. inputthe first line of input is a number T which indicat E The number of cases. Each case contains three lines. The first line is the 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 is in the range of [ -1000000, 1000000]. outputfor, should, and output one line which only Contain K described above. If No such K exists, output-1 instead. sample Input213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 5 1 2 1 2 3 1 2 3 1 3 2 1 2 3 2 1 sample output6-1 &nbspThe  //low method should be 1.1 points, but the use of KMP algorithm will be very cool. (KMP can also be used in the array!!) (⊙o⊙)//The problem is very interesting, let me know the algorithm is more important is the idea, do not have to limit their own thinking   

#include <iostream>#include<cstdio>using namespacestd;intdata1[1000003],data2[1000003];intnex[1000003],n,m;voidGetNext () {intI=0, j=-1; nex[0]=-1;  while(i<m) {if(j==-1|| data2[i]==Data2[j]) {i++;j++; if(data2[i]==Data2[j]) nex[i]=Nex[j]; ElseNex[i]=J; }        ElseJ=Nex[j]; }}intKMP () {intI=0, j=0;    GetNext ();  while(i<N) {if(j==-1|| data1[i]==Data2[j]) {i++;j++; }        Else{J=NEX[J];//J=nex[i] Here again remember the wrong ....         }        if(j==m)returni-m+1; }    return-1;}intMain () {intl; CIN>>l;  while(l--) {cin>>n>>m;  for(intI=0; i<n;i++) scanf ("%d",&Data1[i]);  for(intI=0; i<m;i++) scanf ("%d",&Data2[i]);        GetNext (); intans=KMP (); cout<<ans<<Endl; }    return 0;}

Hdu1711:number Sequence