hdu-5918 Sequence I(kmp)

Posted 2020-08-13 LittlePointer

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

Problem Description

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.

 

 

Input

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).

 

 

Output

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.

 

 

Sample Input

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

 

 

Sample Output

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;
}