• ~1^3 + 2^3 + ... + n^3 = ( \frac{n\times (n+1)}{2} )^2~

• ~1 + 3 + ... + (2n-1)=n^2~

Code mẫu

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;

template <class T>
void read(T &x)
{
    x = 0;
    register int c;
    while ((c = getchar()) && (c > '9' || c < '0'))
        ;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
}

constexpr bool typetest = 0;
constexpr int N = 1e6 + 5;
constexpr int Inf = 1e9 + 7;

int n;

void Read()
{
    cin >> n;
}

void Solve()
{
    for(int i = 1; i <= n; ++i)
        cout << i * i * i << " ";
}

int32_t main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    if (fopen("tests.inp", "r"))
    {
        freopen("test.inp", "r", stdin);
        freopen("test.out", "w", stdout);
    }

    int t(1);
    if (typetest)
        cin >> t;

    for (int _ = 1; _ <= t; ++_)
    {
        // cout << "Case #" << _ << endl;
        Read();
        Solve();
    }
    // cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}

/*
Input:
1
abcahsdfaw
1
q i

Output:
9
*/