hdu1711 Number Sequence

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a
, 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
.
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

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

非常久没有做kmp的题了，这几天要做一下。这是道模板题，直接套模板即可。

#include<stdio.h>
#include<string.h>
int s1[1000005],s2[10006],next[10006],n,m;
void nextt()
{
int i,j;
i=0;j=-1;
memset(next,-1,sizeof(next));
while(i<m){
if(j==-1 || s2[i]==s2[j]){
i++;j++;next[i]=j;
}
else j=next[j];
}
}

int kmp()
{
int i,j;
i=0;j=0;
while(i<n && j<m){
if(j==-1 || s1[i]==s2[j]){
i++;j++;
}
else j=next[j];
}
if(j>=m){
return i+1-m;
}
else return 0;
}

int main()
{
int i,j,T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++){
scanf("%d",&s1[i]);
}
for(i=0;i<m;i++){
scanf("%d",&s2[i]);
}
nextt();
if(kmp()){
printf("%d\n",kmp());
}
else printf("-1\n");
}
return 0;
}