LOJ 6241. 性能优化 I 题解

题给代码意为

\[\begin{align*} &\sum\limits_{i = 1}^n \sum\limits_{j = 1}^{\lfloor \frac n i \rfloor} \sum\limits_{k = 1}^j [\gcd(j, k) = 1] \\ = &\sum\limits_{i = 1}^n \sum\limits_{j = 1}^{\lfloor \frac n i \rfloor} \varphi(j) \\ \end{align*} \]

设

\[S(n) = \sum\limits_{i = 1}^n \varphi(i) \]

则答案为

\[\sum\limits_{i = 1}^n S \left( \left\lfloor \frac n i \right\rfloor \right) \]

发现它很像杜教筛式子

\[S(n) = \frac 1 2 n (n + 1) - \sum\limits_{i = 2}^n S \left( \left\lfloor \frac n i \right\rfloor \right) \]

所以直接代入，答案为

\[\begin{align*} &\sum\limits_{i = 1}^n S \left( \left\lfloor \frac n i \right\rfloor \right) \\ = &S(n) + \sum\limits_{i = 2}^n S \left( \left\lfloor \frac n i \right\rfloor \right) \\ = &\frac 1 2 n (n + 1) - \sum\limits_{i = 2}^n S \left( \left\lfloor \frac n i \right\rfloor \right) + \sum\limits_{i = 2}^n S \left( \left\lfloor \frac n i \right\rfloor \right) \\ = &\frac 1 2 n (n + 1) \\ \end{align*} \]

时间复杂度 \(O(\lg n)\)。

#include <iostream>

#define int long long

using namespace std;

const int mod = 998244353;
const int _2 = 499122177;

static inline int read() {
    int x = 0;
    char ch = cin.get();
    while (!isdigit(ch))
        ch = cin.get();
    while (isdigit(ch)) {
        x = ((x << 3) + (x << 1) + (ch ^ 48)) % mod;
        ch = cin.get();
    }
    return x;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        int n = read();
        cout << n * (n + 1) % mod * _2 % mod << endl;
    }
    return 0;
}