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()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 410;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int a[55][55], f[55], dp[55][55][1305];

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, m;
    while (cin >> n >> m) {
        for (int i = 1; i <= n; i++) {
            f[i] = f[i - 1] + i;
            for (int j = 1; j <= n - i + 1; j++) {
                cin >> a[i][j];
                a[i][j] += a[i - 1][j];
            }
        }
        mem(dp, 0);
        int ans = 0;
        for (int i = n; i; i--)
            for (int j = 0; j <= n - i + 1; j++){
                for (int k = f[j]; k <= m; k++)
                    for (int p = max(j - 1, 0); p < n - i + 1; p++) {
                        dp[i][j][k] = max(dp[i][j][k], dp[i + 1][p][k - j] + a[j][i]);

                    }
                    ans = max(ans, dp[i][j][m]);
            }
        cout << ans << endl;
    }

    return 0;
}