[SOJ #112]Dirichlet 前缀和

题目大意：给定一个长度为$n$的序列$a_n$，需要求出一个序列$b_n$，满足：
$$
b_k=sumlimits_{i|k}a_i
$$
$nleqslant10^7$

题解：$mathrm{Dirichlet}$前缀和，考虑把$k$写成一个无穷向量$[eta_1,eta_2,eta_3,cdots]$，满足$k=sumlimits_iP_i^{eta_i}$，$P_i$为第$i$个质数。相同的，把$i$写成$[alpha_1,alpha_2,alpha_3,cdots]$，于是：

$$
egin{align*}
b_{eta_{k,1},eta_{k,2},eta_{k,2},cdots}&=sumlimits_{i|k}a_{alpha_{i,1},alpha_{i,2},alpha_{i,3},cdots}\
&=sumlimits_{foralleta_{k,j}geqslantalpha_{i,j}}a_{alpha_{i,1},alpha_{i,2},alpha_{i,3},cdots}
end{align*}
$$
于是先线性筛出质数，再做一个高维前缀和即可。复杂度$O(nloglog n)$

卡点：无

C++ Code：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
const int maxn = 1e7 + 5;
typedef unsigned int uint;
typedef unsigned long long ull;

struct random {
	ull seed;
	static const ull multiplier = 0x5deece66dll;
	static const ull addend = 0xbll;
	static const ull mask = 0xffffffffffffll;

	void set_seed (ull _seed) {seed = _seed;}

	uint next() {
		seed = (seed * multiplier + addend) & mask;
		return seed >> 16;
	}
} rnd;

int plist[maxn >> 3], ptot;
bool notp[maxn];
void sieve(const int n) {
	for (int i = 2; i <= n; ++i) {
		if (!notp[i]) plist[ptot++] = i;
		for (int j = 0, t; (t = i * plist[j]) <= n; ++j) {
			notp[t] = 1;
			if (i % plist[j] == 0) break;
		}
	}
}

int n;
uint s[maxn];
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> n >> rnd.seed;
	for (int i = 1; i <= n; ++i) s[i] = rnd.next();
	sieve(n);
	for (int i = 0; i < ptot; ++i) {
		const int P = plist[i];
		for (int j = 1, t; (t = j * P) <= n; ++j)
			s[t] += s[j];
	}
	uint ans = 0;
	for (int i = 1; i <= n; ++i) ans ^= s[i];
	std::cout << ans << '
';
	return 0;
}