算法1

第一列纯1，特判1即可
第二列d=1等差数列，行数是n+1
第三列等差数列的和，不偷懒了，上个二分查找

其后第四列起均为C(3+,n)，行数很少，枚举即可

时间复杂度

二分等差数列和 O(loglogn)

参考文献

C++ 代码

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

ll qs(ll n) {
    if (n < 2)return INT_MAX;
    return n * (n + 1) / 2;
}

ll com(ll u, ll d) {
    if (u > d)return -1ll;
    ll ans = 1;
    for (ll i = 1; i <= u; i++) {
        ans *= d - i + 1;
        ans /= i;
    }
    return ans;
}

ll cq(ll a, ll b) {
    return qs(a) + b;
}

int main() {    
    ll n;
    cin >> n;
    ll ans = cq(n, 2);
    if (n == 1) {
        cout << 1;
        return 0;
    }

    ll l = 4, r = INT_MAX;
    while (l < r) {
        ll m = (l + r) / 2;
        ll t = qs(m - 1);
        if (t < n) {
            l = m + 1;
        }
        else if (t > n) {
            r = m - 1;
        }
        else {
            ans = min(ans, cq(m, 3));
            break;
        }
    }

    for (ll u = 4; u < 1e3; u++) {
        for (ll d = u * 2; 1; d++){
            ll t = com(u - 1, d);
            if (t > n)break;
            if (t == n) {
                ans = min(ans, cq(d, u));
                break;
            }
        }
    }

    cout << ans;
    return 0;
}