LOJ 6241. 性能优化 I 题解

题给代码意为

$$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{k=1}^j[gcd(j,k)=1]\\ =& \sum _{i=1}^n\sum _{j=1}^{\lfloor \frac{n}{i}\rfloor }\phi (j)\end{array}$$

设

$$S(n)=\sum _{i=1}^n\phi (i)$$

则答案为

$$\sum _{i=1}^{n}S\left(\lfloor \frac{n}{i}\rfloor \right)$$

发现它很像杜教筛式子

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

所以直接代入，答案为

$$\begin{array}{rl}& \sum _{i=1}^{n}S\left(\lfloor \frac{n}{i}\rfloor \right)\\ =& S(n)+\sum _{i=2}^{n}S\left(\lfloor \frac{n}{i}\rfloor \right)\\ =& \frac{1}{2}n(n+1)-\sum _{i=2}^{n}S\left(\lfloor \frac{n}{i}\rfloor \right)+\sum _{i=2}^{n}S\left(\lfloor \frac{n}{i}\rfloor \right)\\ =& \frac{1}{2}n(n+1)\end{array}$$

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;
}