[洛谷P4213]【模板】杜教筛（Sum）

题目大意：给你$n$，求：
$$
sumlimits_{i=1}^nvarphi(i),sumlimits_{i=1}^nmu(i)
$$
最多$10$组数据，$nleqslant2^{31}-1$

题解：杜教筛，用来求$sumlimits_{i=1}^nf(i)$的，其中$f$是某个特殊函数。

若我们可以找到一个函数$g$，使得$g,f*g$两个函数的前缀和十分好算（$g*f$表示$g$和$f$的狄利克雷卷积），就可在$O(n^{frac 23})$的复杂度内求出我们要的东西。令$S(n)=sumlimits_{i=1}^nf(i)$
$$
egin{align*}
sumlimits_{i=1}^n(g*f)(i)&=sumlimits_{i=1}^nsumlimits_{d|i}g(d)fleft(dfrac id ight)\
&=sumlimits_{d=1}^ng(d)sumlimits_{i=1,d|i}^nfleft(dfrac id ight)\
&=sumlimits_{d=1}g(d)Sleft(leftlfloordfrac nd ight floor ight)
end{align*}\
g(1)S(n)=sumlimits_{i=1}^n(f*g)(i)-sumlimits_{i=2}^ng(i)Sleft(leftlfloordfrac ni ight floor ight)
$$
然后线性筛求出前$n^{frac23}$项，剩余的递归+整除分块计算，需要记忆化。

$$
sumlimits_{i=1}^nvarphi(i)：\
ecausevarphi*I=id\
herefore S(n)=frac{n(n+1)}{2}-sum_{i=2}^nSleft(leftlfloordfrac n i ight floor ight)
$$
$$
sumlimits_{i=1}^nmu(i)：\
ecausemu*I=1\
herefore S(n)=1-sum_{i=2}^nSleft(leftlfloorfrac n i ight floor ight)
$$

注意这道题中$nleqslant2^{31}-1$，$+1$后会爆$int$，所以整除分块的时候要用$unsigned$

卡点：无

C++ Code：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <unordered_map>
const int R = 1 << 22 | 1;
int plist[R >> 3], ptot;
bool notp[R];
long long phi[R], mu[R];
void sieve() {
	phi[1] = mu[1] = 1;
	for (int i = 2; i < R; ++i){
		if (!notp[i]) plist[ptot++]=i, phi[i] = i - 1, mu[i] = -1;
		for(int j = 0, t; (t = i * plist[j]) < R; ++j){
			notp[t] = true;
			if(i % plist[j] == 0) {
				phi[t] = phi[i] * plist[j];
				mu[t] = 0;
				break; }
			phi[t] = phi[i] * phi[plist[j]];
			mu[t] = -mu[i];
		}
	}
	for (int i = 2; i < R; ++i) phi[i] += phi[i - 1], mu[i] += mu[i - 1];
}
std::unordered_map<unsigned, long long> Phi, Mu;
long long sphi(unsigned n) {
	if (n < R) return phi[n];
	if (Phi.count(n)) return Phi[n];
	long long &t = Phi[n];
	for (unsigned l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		t += (r - l + 1) * sphi(n / l);
	}
	return t = static_cast<long long> (n + 1) * n / 2 - t;
}
long long smu(unsigned n) {
	if (n < R) return mu[n];
	if (Mu.count(n)) return Mu[n];
	long long &t = Mu[n];
	for (unsigned l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		t += (r - l + 1) * smu(n / l);
	}
	return t = 1 - t;
}

int T;
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cin.tie(0);
	sieve();
	std::cin >> T;
	while (T --> 0) {
		static unsigned n;
		std::cin >> n;
		std::cout << sphi(n) << ' ' << smu(n) << '
';
	}
	return 0;
}