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Problem J: Arithmetic Sequence

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 1804 Solved: 308

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Description

Giving a number sequence A with
length n, you should choosing m numbers from A(ignore
the order) which can form an arithmetic sequence and make m as large as possible.

Input

There are multiple test cases. In each test case, the first line contains a positive
integer n. The second line contains n integers
separated by spaces, indicating the number sequence A. All the integers are positive and not more than 2000. The input will end by EOF.

Output

For each test case, output the maximum as
the answer in one line.

Sample Input

5
1 3 5 7 10
8
4 2 7 11 3 1 9 5

Sample Output

4
6

HINT

In the first test case, you should choose 1,3,5,7 to form the arithmetic sequence and its length is 4.

In the second test case, you should choose 1,3,5,7,9,11 and the length is 6.

#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include<algorithm>

#include<vector>

#include<map>

#include<queue>

#include<set>

using namespace std;

const int N=5e4+9;

#define pb push_back

int a[100001];

int q[100001];

int haha[100001];

int main()

{

int n;

while(scanf("%d",&n)==1)

{

memset(q,0,sizeof(q));

memset(haha,0,sizeof(haha));

int maxa=-10;

int maxaa=-10;

for(int i=1;i<=n;i++)

{

cin>>a[i];

haha[a[i]]++;

}

sort(a+1,a+1+n);

for(int i=1;i<=n;i++)

{

if(haha[a[i]]>maxaa)

{

maxaa=haha[a[i]];

}

}

for(int i=1;i<=a
-a[1];i++)

{

int l=1,r=1;

while(l<=n&&r<=n)

{

if(a[r]-a[l]==i)

{

q[i]++;

l=r;

r++;

}

else

{

if(a[r]-a[l]>i)

{

if(maxa<q[i])

{

maxa=q[i];

}

l++;

r--;

q[i]=0;

}

r++;

}

}

if(maxa<q[i])

{

maxa=q[i];

}

}

if(maxa+1>maxaa)

cout<<maxa+1<<endl;

else

cout<<maxaa<<endl;

}

return 0;

}

/*7

1 2 3 4 6 8 10

*/

#include<iostream>

#include<algorithm>

#include<stdio.h>

#include<set>

#include<vector>

#include<string.h>

#include<map>

const int N=1e4+5;

using namespace std;

int s[3000];

int c[3000];

int d[3000];

int e[3000];

int main()

{

int n;

while(scanf("%d",&n)!=EOF)

{

memset(c,0,sizeof(c));

memset(d,0,sizeof(d));

int a;

for(int i=0;i<n;i++)

{

scanf("%d",&a);

c[a]++;

}

int l=0;

int m=0;

for(int i=0;i<=2000;i++)

{

if(c[i]>m) m=c[i];

if(c[i]) s[l++]=i;

}

for(int i=0;i<l;i++)

{

for(int j=i+1;j<l;j++)

{

int x=s[j]-s[i];

if(x<0)d[-x]=1;

else d[x]=1;

}

}

l=0;

for(int i=0;i<=2000;i++)

{

if(d[i]) e[l++]=i;

}

for(int j=0;j<l;j++)

{

for(int i=1;i<=2000;i++)

{

int cnt=1;

int k=i;

if(c[k])

{

while(c[k+e[j]]&&k+e[j]<2000)

{

cnt++;

k=k+e[j];

}

if(cnt>m)

{

m=cnt;

}

}

}

}

cout<<m<<endl;

}

return 0;

}

一个个枚举最好，我是想不出来吗