hdu-5918 Sequence I(kmp)

题目链接：

Sequence I

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 687 Accepted Submission(s): 262


[align=left]Problem Description[/align]
Mr. Frog has two sequences a1,a2,⋯,an and b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bm is exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤n and q≥1.

[align=left]Input[/align]
The first line contains only one integer T≤100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106.

The second line contains n integers a1,a2,⋯,an(1≤ai≤109).

the third line contains m integers b1,b2,⋯,bm(1≤bi≤109).

[align=left]Output[/align]
For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

[align=left]Sample Input[/align]

2
6 3 1
1 2 3 1 2 3
1 2 3
6 3 2
1 3 2 2 3 1
1 2 3

[align=left]Sample Output[/align]

Case #1: 2

Case #2: 1

题意：

问a数组里面匹配的b有多少个;

思路：

把a里面间隔p的拿出来,然后用b数组跑kmp求;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
int n,m,p,a[maxn],b[maxn],c[maxn],nex[maxn];
inline void makenext()
{
int k=-1,j=0;nex[0]=-1;
while(j<m)
{
if(k==-1||b[j]==b[k])
{
j++;k++;
nex[j]=k;
}
else k=nex[k];
}
}
inline int kmp(int l)
{
int pb=0,pc=0,ans=0;
while(pb<m&&pc<l)
{
if(pb==-1||b[pb]==c[pc])pb++,pc++;
else pb=nex[pb];
if(pb==m)ans++,pb=nex[pb];
}
return ans;
}
int main()
{
int t,Case=0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&p);
for(int i=0;i<n;i++)scanf("%d",&a[i]);
for(int i=0;i<m;i++)scanf("%d",&b[i]);
makenext();
int ans=0;
for(int i=0;i<p;i++)
{
int cnt=0;
for(int j=i;j<n;j+=p)c[cnt++]=a[j];
ans+=kmp(cnt);
}
printf("Case #%d: %d\n",++Case,ans);
}
return 0;
}