\[\color{red}{\text{校長者，真神人也，左馬桶，右永神，會執利筆破邪炁，何人當之？}} \\ \begin{array}{|} \hline \color{pink}{\text{The principal is really a god}} \\ \color{pink}{\text{with a closestool on the left and Yongshen on the right}} \\ \color{pink}{\text{holding a sharp pen to pierce the truth}} \\ \color{pink}{\text{Who can resist him? }} \\ \hline \end{array} \\ \begin{array}{|} \hline \color{green}{\text{校長は本當に神であり、左側にトイレ、右側にヨンシェンがあり}} \\ \color{green}{\text{鋭いペンを持って真実を突き刺している。誰が彼に抵抗できるだろうか？ }} \\ \hline \end{array} \\ \begin{array}{|} \hline \color{lightblue}{\text{Le principal est vraiment un dieu}} \\ \color{lightblue}{\text{avec des toilettes à gauche et Yongshen à droite}} \\ \color{lightblue}{\text{tenant un stylo pointu pour percer la vérité}} \\ \color{lightblue}{\text{Qui peut lui résister ? }} \\ \hline \end{array} \\ \begin{array}{|} \hline \color{purple}{\text{Der Direktor ist wirklich ein Gott}} \\ \color{purple}{\text{mit einer Toilette links und Yongshen rechts}} \\ \color{purple}{\text{der einen spitzen Stift hält}} \\ \color{purple}{\text{um die Wahrheit zu durchdringen.}} \\ \color{purple}{\text{Wer kann ihm widerstehen? }} \\ \hline \end{array} \\ \begin{array}{|} \hline \color{cyan}{\text{Principalis deus est, Yongshen a dextris cum latrina}} \\ \color{cyan}{\text{acuto stylo ad perforandum veritatem: quis resistet ei? }} \\ \hline \end{array} \\ \color{red}{\text{對曰：“無人，狗欲當之，還請賜教！”}} \\ \newcommand\bra[1]{\left({#1}\right)} \newcommand\Bra[1]{\left\{{#1}\right\}} \newcommand\dx[0]{\text{dx}} \newcommand\string[2]{\genfrac{\{}{\}}{0pt}{}{#1}{#2}} \newcommand\down[2]{{#1}^{\underline{#2}}} \newcommand\ddiv[2]{\left\lfloor\frac{#1}{#2}\right\rfloor} \newcommand\udiv[2]{\left\lceil\frac{#1}{#2}\right\rceil} \newcommand\lcm[0]{\operatorname{lcm}} \newcommand\set[1]{\left\{{#1}\right\}} \newcommand\ceil[1]{\left\lceil{#1}\right\rceil} \newcommand\floor[1]{\left\lfloor{#1}\right\rfloor} \]

---

壹、題目描述 ¶

  定義 \(f(x)=\mu^2(x)x\)，求

\[\sum_{i=1}^n f(i) \]

  資料範圍保證 \(n\le 10^{14}\).

貳、題解 ¶

  考慮用小的數去篩大的數。顯然我們只需要用到 \(\sqrt n\) 以內的數了。設 \(S(i)=1+2+3+\cdots+i=\frac{i(i+1)}{2}\)，利用 \(\mu\) 自身的容斥性就可以算出來答案是：

\[\begin{aligned} \sum_{i=1}^{\sqrt n}\mu(i)S\bra{\ddiv{n}{i^2}}\cdot i^2 \end{aligned} \]

  然後就完了，我他媽是大傻逼。時間複雜度可能在 \(\mathcal O(\sqrt n+T\sqrt[3]n)\)

.

叄、參考程式碼 ¶

# define your short
# define coffin vector<your>()
#include <bits/stdc++.h>
using namespace std;

namespace Elaina {

#define rep(i, l, r) for (int i = l, i##_end_ = r; i <= i##_end_; ++i)
#define drep(i, l, r) for (int i = l, i##_end_ = r; i >= i##_end_; --i)
#define fi first
#define se second
#define Endl putchar('\n')

    template<class T> inline T fab(T x) { return x < 0? -x: x; }
    template<class T> inline void getmax(T& x, const T& rhs) { x = max(x, rhs); }
    template<class T> inline void getmin(T& x, const T& rhs) { x = min(x, rhs); }

    typedef unsigned usint;
    typedef long long ll;
    typedef unsigned long long ull;

} // namespace Elaina
using namespace Elaina;

const int maxn = 1e7;

int mu[maxn + 5], prime[maxn + 5], pcnt;
ull pre[maxn + 5];
bool sie[maxn + 5];
inline void sieve() {
    mu[1] = 1, sie[1] = true;
    for (int i = 2; i <= maxn; ++i) {
        if (!sie[i]) prime[++pcnt] = i, mu[i] = -1;
        for (int j = 1; j <= pcnt && i * prime[j] <= maxn; ++j) {
            sie[i * prime[j]] = true;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0; break;
            }
            else mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= maxn; ++i)
        pre[i] = pre[i - 1] + (ull)i * i * mu[i];
}
inline ull S(ll x) {
    ull ret;
    if (x & 1) ret = (ull)((x + 1) >> 1) * x;
    else ret = (ull)(x >> 1) * (x + 1);
    return ret;
}
signed main() {
    cin.tie(NULL) -> sync_with_stdio(false);
    sieve();
    int _; cin >> _; ll n;
    while (_--) {
        cin >> n;
        ull ans = 0;
        for (ll l = 1, r; l * l <= n; l = r + 1) {
            r = sqrt(n / (n / l / l));
            if(n / r / r == n / l / l) ++r;
            if(n / r / r != n / l / l) --r;
            ans += 1ll * (pre[r] - pre[l - 1]) * S(n / l / l);
        }
        printf("%llu\n", ans);
    }
    return 0;
}