1 条题解

  • 1
    @ 2024-11-27 8:39:36
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int N = 110, INF = 0x3f3f3f3f, MOD = 1E9 + 7;
    int n, a[N][N], f[N][N];
    
    int main(int argc, char* argv[]) {
        cin >> n;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= i; j++)
                cin >> a[i][j];
    
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= i; j++)
                f[i][j] = max(f[i - 1][j - 1], f[i - 1][j]) + a[i][j];
    
        int ans = -2e9;
        for (int i = 1; i <= n; i++)
            ans = max(ans, f[n][i]);
        cout << ans;
        return 0;
    }
    
    • 1

    信息

    ID
    1135
    时间
    1000ms
    内存
    512MiB
    难度
    1
    标签
    递交数
    105
    已通过
    90
    上传者