[洛谷P4900]食堂

题目大意：$n(nleqslant10^6)$组询问，每组询问给出$l,r(l,rleqslant10^6)$，求（${dfrac ij}$表示$dfrac ij$的小数部分）：

$$
sumlimits_{i=l}^rsumlimits_{j=1}^i{dfrac ij}pmod{998244353}
$$

题解：令$f(x)=sumlimits_{i=1}^x{dfrac xi}$
$$
f(x+1)=sumlimits_{i=1}^{x+1}{dfrac{x+1}i}\
f(x+1)=sumlimits_{i=1}^{x+1}{dfrac xi+dfrac 1i}\
ecause {dfrac{x+1}{x+1}}=0\
herefore f(x+1)=f(x)+sumlimits_{i=1}^{x}{dfrac 1i}-sumlimits_{i=1}^{x+1}[{dfrac{x+1}i}=0]+1\
令s_0(x)为x的约数个数\
即f(x+1)=f(x)+sumlimits_{i=1}^{x}{dfrac 1i}-s_0(x+1)+1
$$
求几次前缀和即可

卡点：无

C++ Code：

#include <cstdio>
#include <iostream>
#define maxn 1000005
const int mod = 998244353;
inline void reduce(int &x) { x += x >> 31 & mod; }
inline int getreduce(int x) { return x + (x >> 31 & mod); }

int n;
int inv[maxn], sinv[maxn], f[maxn];
int ans[maxn], sans[maxn];
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	inv[1] = sinv[1] = 1;
	for (int i = 2; i < maxn; ++i) {
		inv[i] = static_cast<long long> (mod - mod / i) * inv[mod % i] % mod;
		reduce(sinv[i] = sinv[i - 1] + inv[i] - mod);
	}
	for (int i = 2; i < maxn; ++i)
		for (int j = i; j < maxn; j += i) ++f[j];
	for (int i = 2; i < maxn; ++i) {
		reduce(ans[i] = ans[i - 1] + sinv[i - 1] - mod);
		reduce(ans[i] -= f[i]);
		reduce(sans[i] = sans[i - 1] + ans[i] - mod);
	}
	std::cin >> n;
	while (n --> 0) {
		static int l, r;
		std::cin >> l >> r;
		std::cout << getreduce(sans[r] - sans[l - 1]) << '
';
	}
	return 0;
}