51Nod

给出N个正整数，要求选出一些数组成等差数列。

问最长的等差数列的长度是多少。

3 <= N <= 10000  2<= Ai <= 1e9

容易想到用DP的方法。

dp[i][j]表示以j为中间的项拓展的最大的等差数列长度。

因此可以用双指针拓展，复杂度N*N。这里用short存储可以避免MLE

#pragma warning(disable:4996)

#include<iostream>
#include<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_map>
#include<fstream>
#include<iomanip>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define inf 0x7FFFFFFF
#define MOD 998244353
#define moD 1000000003
#define pii pair<int,string>
#define eps 1e-8
#define equals(a,b) (fabs(a-b)<eps)
#define bug puts("bug")
#define re  register
#define fi first
#define se second
const int maxn = 1e4 + 5;
const double Inf = 10000.0;
const double PI = acos(-1.0);
typedef  long long ll;
typedef unsigned long long ull;
using namespace std;

short dp[maxn][maxn];
int a[maxn];

int main() {
    int n;
    short res = 2;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", a + i);
    sort(a, a + n);
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            dp[i][j] = 2;
        }
    }
    for (int j = n - 2; j >= 1; j--) {
        int i = j - 1, k = j + 1;
        while (i >= 0 && k < n) {
            if (a[i] + a[k] < (a[j] << 1)) k++;
            else if (a[i] + a[k] > (a[j] << 1)) i--;
            else {
                dp[i][j] = dp[j][k] + 1;
                res = max(res, dp[i][j]);
                i--, k++;
            }
        }
    }
    printf("%d", res);
}