Hangdian 1711 Number Sequence

Number Sequence Time Limit: 10000/5000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 10239 accepted submission (s): 4656


Problem descriptiongiven 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.
 
Inputthe 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 [-1000000,100 0000].
 
Outputfor each test case, You shoshould 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

Classic KMP.

The AC code is as follows:

#include<iostream>#include<cstdio>using namespace std;int a[1000005],b[10005],next[10005];int main(){    int t;    int n,m,ans;    int i,j;    scanf("%d",&t);    while(t--)    {        ans=-1;        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]);        i=0;j=-1;next[0]=-1;        while(i<m)        {            if(j==-1||b[i]==b[j])                next[++i]=++j;            else                j=next[j];        }        i=0;j=0;        while(i<n&&j<m)        {            if(j==-1||a[i]==b[j])            i++,j++;            else j=next[j];            if(j==m)            {                ans=i-j+1;break;            }        }        printf("%d\n",ans);    }    return 0;}

Hangdian 1711 Number Sequence