Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8600 Accepted Submission(s): 3953
Problem Description
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b (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. 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. 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
Source
Recommend
lcy
没事水一把KMP。。。。。
#include <iostream> #include <cstdio> #include <cstring>
using namespace std;
int n,m;
int a[1000010],b[1000010],next[1000010];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(next,0,sizeof(next));
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
}
for(int i=0;i<m;i++)
{
scanf("%d",b+i);
}
for(int i=1;i<m;i++)
{
int j=i;
while(j>0)
{
j=next[j];
if(b==b[j])
{
next[i+1]=j+1;
break;
}
}
}
int pos=-2;
for(int i=0,j=0;i<n;i++)
{
if(j<m&&a==b[j])
{
j++;
}
else
{
while(j>0)
{
j=next[j];
if(a==b[j])
{
j++;
break;
}
}
}
if(j==m)
{
pos=i-m+1;
break;
}
}
printf("%d\n",pos+1);
}
return 0;
}
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* This source code was highlighted by YcdoiT. ( style: Codeblocks )
