题解 P4900 食堂

一道推式子的数学题。

\begin{aligned} \sum_{i = l}^{r} \sum_{j = 1}^{i} {\frac{j}{i}} & = \sum_{i = l}^{r} \sum_{j = 1}^{i} (\frac{j}{i} - ⌊ \frac{j}{i} ⌋) \\ = \sum_{i = l}^{r} \sum_{j = 1}^{i} \frac{j}{i} - \sum_{i = l}^{r} \sum_{j = 1}^{i} ⌊ \frac{j}{i} ⌋ \end{aligned}

$\begin{aligned} \sum\limits_{i=l}^{r}\sum\limits_{j=1}^{i}\left\{\dfrac{j}{i}\right\} &=\sum\limits_{i=l}^{r}\sum\limits_{j=1}^{i}\left(\dfrac{j}{i}-\left\lfloor\dfrac{j}{i}\right\rfloor\right)\\ &={\sum\limits_{i=l}^{r}\sum\limits_{j=1}^{i}}\dfrac{j}{i}- {\sum\limits_{i=l}^{r}\sum\limits_{j=1}^{i}}\left\lfloor\dfrac{j}{i}\right\rfloor \end{aligned}$

令 $g(x)$ 为 $1$ 到 $x$ 的所有数的约数个数之和。即： $\sum\limits_{i=1}^{x}\dfrac{x}{i}$

则最终式子可以化简为： $\sum\limits_{i=l}^{r}\left[i\cdot\left(\sum\limits_{j=1}^{i}j^{-1}\right)\right]-\sum\limits_{i=l}^{r}g(i)$

第一项和第二项均可用两次前缀和预处理。

预处理 $O(n\log n)$ ，查询 $O(1)$ 。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5 + 5, N2 = 1e6 + 5, mod = 998244353;
ll T, s = 0, inv[N2], cnt[N2], x[N2], f[N2];
ll getans(ll n) {return (x[n] - f[n] + mod) % mod;}
int main() {
    inv[1] = s = x[1] = f[1] = cnt[1] = 1;
    for (ll i = 2; i <= 1e6; ++i) {
    	inv[i] = (mod - mod / i * inv[mod % i] % mod) % mod;
		s = (s + inv[i]) % mod;
		x[i] = (x[i - 1] + i * s) % mod;
	}
    for (ll i = 2; i <= 1e6; ++i) {
        ++cnt[i];
        for (int j = i; j <= 1e6; j += i) ++cnt[j];
    }
    s = 0;
    for (int i = 1; i <= 1e6; ++i) s = (s + cnt[i]) % mod, f[i] = (f[i - 1] + s) % mod;
    scanf("%d", &T); 
    while (T--) {
        ll p, q; scanf("%lld%lld", &p, &q);
        printf("%lld\n", (getans(q) - getans(p - 1) + mod) % mod);
    }
    return 0;
}