题目链接

题意：



题解：
















AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9+7;
//const int mod = 998244353;
const double eps = 1e-10;
const double pi = acos(-1.0);
const int maxn = 1e6+10;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const int N = 8010;
ll dp[N], f[N][N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j < i; j++)
            dp[i] = max(dp[i], dp[j] + f[i - j - 1][j] + i - j - 1);
        for (int j = 1; j <= n; j++)
            f[j][i] = max(f[j][i - 1], (j >= i? f[j - i][i] + dp[i] : 0));
    }
    cout << dp[n] << endl;
    return 0;
}