HDU-5918 Sequence I（暴力）（KMP）

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 422    Accepted Submission(s): 158


[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

 

[align=left]Source[/align]
2016中国大学生程序设计竞赛（长春）-重现赛
  

  开始看见这个题，感觉复杂度挺模糊的，所以想暴力一发，然而自己写挫了，想了下将a数组拆开，感觉太麻烦了，就弃了，结果这两个做法都是可行的，第二种奇妙的可以用KMP，复杂度低多了。所以，有想法就要上！！！

1.暴力，直接枚举q，然后我是判断条件写挫了，看了别人的暴力好久才找到，竟然差点以为while和for有区别。This

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1000005;
int a
,b
;

int main()
{
int t,n,m,p;
cin >> t;
for(int ca = 1;ca <= t;ca++)
{
scanf("%d%d%d",&n,&m,&p);
for(int i = 1;i <= n;i++)
scanf("%d",&a[i]);
for(int i = 1;i <= m;i++)
scanf("%d",&b[i]);
printf("Case #%d: ",ca);
int sum = 0;
for(int i = 1;i <= n && i+(m-1)*p <= n;i++)
{
int j = 0,r = i;
while(a[r] == b[j+1])
{
j++;
r = i + j*p;
if(j >= m)
{
sum++;
break;
}
}
}
printf("%d\n",sum);
}
return 0;
}

2.KMP，可以将数组抽出来，也可以将kmp模板中的起点和++修改一下。

卧槽，貌似我以前写的kmp模板都是错的，，，

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int Next[2000100];
int a[2000100],b[2000100],c[2000100];
void Getfail(int m)
{
int j, k;
j = 0; k = -1; Next[0] = -1;
while(j < m)
if(k == -1 || b[j] == b[k])
Next[++j] = ++k;
else
k = Next[k];

}
int KMP(int n,int m)
{
int i = 0,j = 0,ans = 0;
while(i < n)
{
if(j == -1 || c[i] == b[j])
{
i++,j++;
if(j == m)
{
j = Next[j];
ans++;
}
}
else
j = Next[j];
}
return ans;
}
int main()
{
int t,n,m,p;
cin >> t;
for(int ca = 1;ca <= t;ca++)
{
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]);
printf("Case #%d: ",ca);
int sum = 0;
Getfail(m);
for(int i = 0;i < p;i++)
{
int num = 0;
for(int j = i;j < n && i+(m-1)*p < n;j += p)
c[num++] = a[j];
sum += KMP(num,m);
}
printf("%d\n",sum);
}
return 0;
}