「SuperOJ 1297」数学题-ExtendedEratosthenesSieve

定义欧拉函数的变种
$$\phi(n, d) = \prod_{k = 1} ^ m (p_k ^ {e_k} + d)$$
特别的,$\phi(1, d) = 1$,求
$$\sum_{i = 1} ^ n \phi(i, d) \pmod {10 ^ 9 + 7}$$

题解

首先 $d$ 是给定的值,令 $f(n) = \phi(n, d)$,考虑扩展埃拉托色尼筛法。
$f(1) = 1$
当 $n = p$,$p$ 是质数时,$f(p) = p + d$,
当 $n = p ^ e$ 时,$f(p ^ e) = p ^ e + d$

所以我们需要预处理 $p ^ 0, p ^ 1$ 的前缀和,$f(p)$ 可以通过 $d \times p ^ 0 + p$ 计算,枚举指数时单独维护 $p ^ e$ 来保证复杂度即可。

代码

和筛欧拉函数的代码一样简单(逃)

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/**
* Copyright (c) 2017-2018, xehoth
* All rights reserved.
* 「SuperOJ 1297」数学题 24-01-2018
* Extended Eratosthenes Sieve
* @author xehoth
*/
#include <bits/stdc++.h>
int d, M, n;
const int MOD = 1e9 + 7;
std::vector<int> pre[2], suc[2], primes;
inline int add(const int x, const int v) {
return x + v >= MOD ? x + v - MOD : x + v;
}
inline int dec(const int x, const int v) {
return x - v < 0 ? x - v + MOD : x - v;
}
int rec(int res, int last, int mul) {
int t = dec((res > M ? suc[1][n / res] : pre[1][res]),
pre[1][primes[last] - 1]);
int ret = (unsigned long long)t * mul % MOD;
for (int i = last, p; i < (int)primes.size(); i++) {
p = primes[i];
if (p * p > res) break;
for (int q = p, nres = res, nmul, pk = p;
(unsigned long long)p * q <= res; q *= p) {
nmul = mul * ((unsigned long long)pk + d) % MOD;
ret = add(ret, rec(nres /= p, i + 1, nmul));
pk = (unsigned long long)pk * p % MOD;
ret = add(ret, mul * ((unsigned long long)pk + d) % MOD);
}
}
return ret;
}
inline int solve(const int n) {
M = sqrt(n);
primes.reserve(M);
pre[0].resize(M + 1);
pre[1].resize(M + 1);
suc[0].resize(M + 1);
suc[1].resize(M + 1);
for (int i = 1, t; i <= M; i++) {
pre[0][i] = i - 1;
pre[1][i] = (i * (i + 1ull) / 2 - 1 + MOD) % MOD;
t = n / i;
suc[0][i] = t - 1;
suc[1][i] = (t * (t + 1ull) / 2 - 1 + MOD) % MOD;
}
for (int p = 2, end; p <= M; p++) {
if (pre[0][p] == pre[0][p - 1]) continue;
primes.push_back(p);
const int q = p * p, m = n / p;
const int pcnt = pre[0][p - 1], psum = pre[1][p - 1];
end = std::min(M, n / q);
for (int i = 1, w = M / p; i <= w; i++) {
suc[0][i] = dec(suc[0][i], dec(suc[0][i * p], pcnt));
suc[1][i] =
dec(suc[1][i],
(unsigned long long)dec(suc[1][i * p], psum) * p % MOD);
}
for (int i = M / p + 1; i <= end; i++) {
suc[0][i] = dec(suc[0][i], dec(pre[0][m / i], pcnt));
suc[1][i] =
dec(suc[1][i],
(unsigned long long)dec(pre[1][m / i], psum) * p % MOD);
}
for (int i = M; i >= q; i--) {
pre[0][i] = dec(pre[0][i], dec(pre[0][i / p], pcnt));
pre[1][i] =
dec(pre[1][i],
(unsigned long long)dec(pre[1][i / p], psum) * p % MOD);
}
}
primes.push_back(M + 1);
for (int i = 1; i <= M; i++) {
pre[1][i] = (pre[1][i] + (unsigned long long)d * pre[0][i]) % MOD;
suc[1][i] = (suc[1][i] + (unsigned long long)d * suc[0][i]) % MOD;
}
return n > 1 ? 1 + rec(n, 0, 1) : 1;
}
int main() {
std::cin >> n >> d;
std::cout << solve(n);
}

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