Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12270    Accepted Submission(s): 5600

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

 

Source

HDU 2007-Spring Programming Contest

 

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分析：这是一道简单的字符串匹配问题，运用的是kmp思想。

代码：

 #include<stdio.h>
#include<string.h>
#include<stdlib.h>
int next[1000010];
int n,m;
int a[1000010];
int b[1000010];
void getnext()
{
int i,j;
next[0]=-1;
i=0;j=-1;
while(i<m)
{
if(j==-1||b[i]==b[j])
{
i++;j++;
if(b[i]!=b[j])
{
next[i]=j;
}
else next[i]=next[j];
}
else j=next[j];
}
}
void kmp()
{
int i,j;
i=j=0;
getnext();
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);

}
else
{
printf("-1\n");
 
}

}


int main()
{

int i,j,k,t;
scanf("%d",&k);
while(k--)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(j=0;j<m;j++)
scanf("%d",&b[j]);
   kmp();
}
return 0;
}