标签：print nbsp find cstring ssi return memory exist printf
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24116 Accepted Submission(s):
10232
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
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
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int len1,len2,a[1000005],b[10005],ne[10005];
void getnext()
{
int i,j;
ne[0]=0;ne[1]=0;
for(i=1;i<len2;i++)
{
j=ne[i];
while(j&&b[i]!=b[j])j=ne[j];
if(b[i]==b[j])ne[i+1]=j+1;
else ne[i+1]=0;
}
}
void kmp()
{
int i,j=0;
int flag=0;
for(i=0;i<len1;i++)
{
while(j&&a[i]!=b[j])j=ne[j];
if(a[i]==b[j])j++;
if(j==len2)
{
flag=1;
printf("%d\n",i-len2+2);
break;
}
}
if(!flag)cout<<-1;
}
int main()
{
int T,i;
cin>>T;
while(T--)
{
cin>>len1>>len2;
for(i=0;i<len1;i++)cin>>a[i];
for(i=0;i<len2;i++)cin>>b[i];
getnext();
kmp();
}
return 0;
}
Number Sequence （HDU 1711）
标签：print nbsp find cstring ssi return memory exist printf
原文地址：http://www.cnblogs.com/thmyl/p/6285995.html
