「补档计划」数论

文章目录
  1. 1. Todo
  2. 2. 扩展欧几里得算法
    1. 2.1. 证明
    2. 2.2. 代码
  3. 3. 逆元
    1. 3.1. 费马小定理
    2. 3.2. 扩展欧几里得
    3. 3.3. 递推式
  4. 4. 中国剩余定理
    1. 4.1. 基本形式
    2. 4.2. 扩展形式
    3. 4.3. 代码
  5. 5. BSGS
    1. 5.1. 代码
  6. 6. 类欧几里得
    1. 6.1. 代码
  7. 7. 欧拉函数
    1. 7.1. 一些性质
    2. 7.2. 求 phi
    3. 7.3. 筛 phi
  8. 8. 狄利克雷(Dirichlet)卷积
    1. 8.1. 定义
    2. 8.2. 一些常见积性函数
    3. 8.3. 一些性质
    4. 8.4. 常见卷积
    5. 8.5. 一些变换
    6. 8.6. 预处理 Dirichlet 卷积
  9. 9. 莫比乌斯函数
    1. 9.1. 定义
    2. 9.2. 性质
  10. 10. 莫比乌斯反演
    1. 10.1. 变形
  11. 11. 杜教筛
    1. 11.1. 前置技能
      1. 11.1.1.
      2. 11.1.2.
    2. 11.2. 主要形式
  12. 12. 「HDU 3579」Hello Kiki
    1. 12.1. 链接
    2. 12.2. 题解
    3. 12.3. 代码
  13. 13. 「BZOJ 1101」Zap
    1. 13.1. 链接
    2. 13.2. 题解
    3. 13.3. 代码
  14. 14. 「BZOJ 2820」YY的GCD
    1. 14.1. 链接
    2. 14.2. 题解
    3. 14.3. 代码
  15. 15. 「SDOI 2014」数表
    1. 15.1. 链接
    2. 15.2. 题解
    3. 15.3. 代码
  16. 16. 「51 NOD 1244」莫比乌斯函数之和
    1. 16.1. 链接
    2. 16.2. 题解
    3. 16.3. 代码
  17. 17. 「51 NOD 1239」欧拉函数之和
    1. 17.1. 链接
    2. 17.2. 题解
    3. 17.3. 代码
  18. 18. 「BZOJ 3944」Sum
    1. 18.1. 链接
    2. 18.2. 题解
    3. 18.3. 代码
  19. 19. 「HDU 5608」function
    1. 19.1. 链接
    2. 19.2. 题解
    3. 19.3. 代码

关于数论的总结和专题练习….

Todo

  • 快速幂
  • gcd
  • exgcd
  • 逆元
  • 中国剩余定理
  • 类欧几里德
  • Miller-Rabin
  • Pollard-Rho
  • 数论函数
  • 线性筛
  • 反演
  • BSGS
  • 斯特林数
  • 原根
  • Fast Fourier Transformation
  • Fast Number-Theoretic Transform
  • 杜教筛
  • 洲阁筛

扩展欧几里得算法

对于不完全为 $0$ 的非负整数 $a, b$,$gcd(a, b)$ 表示 $a, b$ 的最大公约数,必然存在整数对 $x, y$,使得 $gcd(a, b) = ax + by$。

证明

不妨设 $a > b$,显然当 $b = 0$, $gcd(a, b) = a$。此时 $x = 1, y = 0$。

当 $ab \neq 0$ 时,设 $ax_1 + by_1 = gcd(a, b)$,$bx_2 + (a \text{ mod } b)y_2 = gcd(b, a \text{ mod } b)$。

根据欧几里得原理有 $gcd(a, b) = gcd(b, a \text{ mod } b)$。

则 $ax_1 + by_1 = bx_2 + (a \text{ mod } b)y_2 = bx_2 + (a - \lfloor \frac {a} {b} \rfloor \cdot b)y_2 = ay_2 + bx_2 - \lfloor \frac {a} {b} \rfloor \cdot by_2$。

所以 $x_1 = y_2, y_1 = x_2 - \lfloor \frac {a} {b} \rfloor \cdot y_2$,接下来不断递归就好了。

代码

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inline void exgcd(long a, long b, long &g, long &x, long &y) {
!b ? (x = 1, y = 0, g = a) : (exgcd(b, a % b, g, y, x), y -= (a / b) * x);
}

逆元

费马小定理

$$a^{p - 1} \equiv \text{ (mod p)}$$

要求 $p$ 为素数,上述公式可变为

$$a \times a^{p - 2} \equiv \text{ (mod p)}$$

由乘法逆元的定义,$a^{p - 2}$ 为 $a$ 的乘法逆元。

用快速幂计算 $a^{p - 2}$,总时间复杂度为 $O(\text{log }a)$

扩展欧几里得

扩展欧几里得(EXGCD)算法可以在 $O(\text{log max(a, b)})$ 的时间内求出关于 $x$,$y$ 的方程

$$ax + by = gcd(a, b)$$

的一组整数解

当 $b$ 为素数时有 $gcd(a, b) = 1$,直接求解就好了。

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inline long inv(const long num) {
register long g, x, y;
exgcd(num, MOD, g, x, y);
return (x % MOD + MOD) % MOD;
}

递推式

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inv[1] = 1;
for (register int i = 2; i <= n; i++)
inv[i] = (long)(mod - mod / i) * inv[mod % i] % mod;

中国剩余定理

基本形式

设正整数 $m_1, m_2, \cdots, m_n$ 两两互质,则同余方程组
$$x \equiv a_1 \ (\text{mod } m_1)$$
$$x \equiv a_2 \ (\text{mod } m_2)$$
$$\cdots$$
$$x \equiv a_n \ (\text{mod } m_n)$$
有整数解,并且在模 $M = m_1 \cdot m_2 \cdot _\cdots \cdot m_n$ 解是唯一的,$x = \sum_{i = 1} ^ n a_iM_iM_i ^ {-1} \text{ mod M}$,其中 $M_i = M \ / \ m_i$,$M_i ^ {-1}$ 为 $M_i$ 模 $m_i$ 的逆元。

扩展形式

基本形式中的 $m_i$ 不再要求互质,此时只能两两求解,$x \equiv a_1 \ (\text{mod } m_1), x \equiv a_2 \ (\text{mod } m_2)$。

代码

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template <typename T>
inline void exgcd(T a, T b, T &g, T &x, T &y) {
!b ? (x = 1, y = 0, g = a) : (exgcd(b, a % b, g, y, x), y -= (a / b) * x);
}
template <typename T>
inline T getInv(const T num, const T mod) {
register T g, x, y;
exgcd(num, mod, g, x, y);
return (x % mod + mod) % mod;
}
template <typename T>
inline T crt(T *a, T *m, const int n) {
register T M = m[0], ans = 0, mi;
for (register int i = 1; i < n; i++) M *= m[i];
for (register int i = 0; i < n; i++)
mi = M / m[i], ans = (ans + a[i] * mi * getInv(mi, m[i])) % M;
return ans < 0 ? ans + M : ans;
}
template <typename T>
inline T excrt(T *a, T *m, int n) {
register T M = m[0], ans = a[0], g, x, y;
for (register int i = 1; i < n; i++) {
exgcd(M, m[i], g, x, y);
if ((a[i] - ans) % g) return -1;
x = (a[i] - ans) / g * x % (m[i] / g);
ans = (ans + x * M) % (M = M / g * m[i]);
}
return ans > 0 ? ans : ans + M;
}

BSGS

求解方程 $A ^ x \equiv B(\text{mod }C)$。

代码

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inline int bsgs(int a, int b, int c) {
register int cnt = 0, g, d = 1;
while ((g = gcd(a, c)) != 1) {
if (b % g) return -1;
cnt++, b /= g, c /= g, d = (long long)d * (a / g) % c;
}
b = (long long)b * inv(d, c) % c;
map.clear();
register int s = sqrt(c), p = 1;
for (register int i = 0; i < s; i++) {
if (p == b) return i + cnt;
map[(long long)p * b % c] = i, p = (long long)p * a % c;
}
register int q = p;
for (register int i = s; i - s + 1 <= c - 1; i += s) {
Map::iterator it = map.find(q);
if (it != map.end()) return i - it->second + cnt;
q = (long long)q * p % c;
}
return -1;
}

类欧几里得

求解形如 $\sum_{x = 0} ^ n \lfloor \frac {ax + b} {c} \rfloor$。

我们令 $S_d(n) = \sum_{x = 0} ^ n x ^ d$。

  1. 若 $a \geq c$,则令 $d = \lfloor \frac {a} {c} \rfloor$,有$$\sum ^ {n}_{x = 0} \lfloor \frac{ax + b}{c} \rfloor = \sum ^ {n}_{x = 0}(\lfloor \frac {(a \% c)x + b}{c}\rfloor + d * x) = d * S_1(n) + \sum_{x = 0} ^ n\lfloor \frac{(a \% c)x + b}{c} \rfloor$$
  2. 若 $b \geq c$,则令 $d = \lfloor \frac {b} {c} \rfloor$,有$$\sum ^ {n}_{x = 0} \lfloor \frac{ax + b}{c} \rfloor = \sum ^ {n}_{x = 0}(\lfloor \frac{ax + (b \% c)}{c}\rfloor + d) = d * S_0(n) + \sum ^ {n}_{x = 0} \lfloor \frac{ax + (b \% c)}{c} \rfloor$$
  3. 若 $a < c, b < c$ 时,令 $M = \lfloor \frac {an + b} {c} \rfloor$,有$$\sum ^ {n}_{x = 0} \lfloor \frac{ax + b}{c} \rfloor = \sum ^ {n}_{x = 0} \sum ^ {M}_{y = 1}[y \leq \lfloor \frac{ax + b}{c} \rfloor]$$ 即$$\sum ^ {n}_{x = 0} \sum ^ {M - 1}_{y = 0}[y + 1 \leq \lfloor \frac{ax + b}{c} \rfloor]$$ 接下来化简 $y + 1 \leq \lfloor \frac {ax + b} {c} \rfloor$,有$$c \times y + c - b \leq a \times x$$ $$c \times y + c - b - 1 < a \times x$$
    $$x > \lfloor \frac {c \times y + c - b - 1} {a} \rfloor$$
    故原式$$\begin {aligned} = &\sum ^ {M - 1}_{y = 0}\sum ^ {n}_{x = 0}[x > \lfloor \frac{c * y + c - b - 1}{a} \rfloor]&\\ = &\sum ^ {M - 1}_{y = 0}(n - \lfloor \frac{c * y + c - b - 1}{a}\rfloor) &\\ = &n * M - \sum ^ {M - 1}_{y = 0} \lfloor \frac{c * y + c - b - 1}{a}\rfloor &\\ \end {aligned}$$

由于 $a < c$ 时 $\lfloor \frac {a} {c} \rfloor = 0$,$b < c$ 时 $\lfloor \frac {b} {c} \rfloor = 0$,所以我们可以将 1, 2 两种情况合并,即令
$$f(a, b, c, n) = \sum_{x = 0} ^ n \lfloor \frac {ax + b} {c} \rfloor$$
那么 $a \geq c$ 或 $b \geq c$ 时,有
$$f(a, b, c, n) = \lfloor \frac {a} {c} \rfloor * S_1(n) + \lfloor \frac {b} {c} \rfloor * S_0(n) + f(a \% c, b \% c, c, n)$$
将 $S_d(n)$ 代入,得
$$f(a, b, c, n) = \lfloor \frac {a} {c} \rfloor * \frac {n * (n + 1)} {2} + \lfloor \frac {b} {c} \rfloor * (n + 1) + f(a \% c, b \% c, c, n)$$
当 $a < c$ 且 $b < c$ 时,有
$$f(a, b, c, n) = n * M - f(c, c - b - 1, a, M - 1)$$
把 $M$ 代入,得
$$f(a, b, c, n) = n * \lfloor \frac {an + b} {c} \rfloor - f(c, c - b - 1, a, \lfloor \frac {an + b} {c} \rfloor - 1)$$
当 $a = 0$ 时,即为该算法的边界,此时
$f(a, b, c, n) = \lfloor \frac {b} {c} \rfloor * (n + 1)$

我们可以发现 $(a, c)$ 会变化为 $(c, a \% c)$,故这个算法被称为类欧几里德算法。

代码

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#define long long long
inline long classEuclid(long a, long b, long c, long n) {
if (a == 0) {
return (b / c) * (n + 1);
} else if (a >= c || b >= c) {
return (a / c) * (n * (n + 1) >> 1) + (b / c) * (n + 1) +
classEuclid(a % c, b % c, c, n);
} else {
return (n * ((a * n + b) / c)) -
classEuclid(c, c - b - 1, a, (a * n + b) / c - 1);
}
}

欧拉函数

欧拉函数 $\varphi (n)$: $\varphi (n)$ 表示 $\left [1, n \right ]$ 中与 $n$ 互质的整数的个数。

通式: $\varphi (n) = \sum_{i = 1}^{n} \left [ gcd(i, n) = 1 \right ] = n \prod_{i = 1}^n \left ( 1 - \frac{1} {p_i} \right )$,其中 $p_i$ 为 $n$ 的所有质因数,$n$ 为不是 $0$ 的整数。

一些性质

  1. $\varphi (1) = 1$。
  2. 若 $n$ 是质数 $p$ 的 $k$ 次幂,则 $\varphi (n) = p^k - p^{k - 1} = (p - 1)p^{k - 1}$。
  3. 欧拉函数是积性函数,但不是完全积性函数。
  4. 当 $n > 1$ 时,$\sum_{i = 1}^{n} i \cdot \left [ gcd(i, n) = 1 \right ] = \frac{n \cdot \varphi(n)} {2}$。
  5. 欧拉定理:对于互质的整数 $a, n$,有 $a ^ {\varphi(n)} \equiv 1 \ (\text{mod n})$,那么 $a ^ x \equiv a ^ {x \text{ mod } \varphi(n)} \ (\text{mod n})$。
  6. 扩展欧拉定理:对于不互质的整数 $a, n$,$a ^ x \equiv a ^ {x \text{ mod } \varphi(n) + \varphi(n)} \ (\text{mod }n)(x \geq \varphi(n))$。
  7. 若 $p \ | \ i$,$p$ 是质数,那么 $\varphi(i * p) = p * \varphi(i)$,否则 $\varphi(p * i) = (p - 1) * \varphi(i)$
  8. $\sum_{d | n}\varphi(d) = n$

求 phi

根据通式暴力求解即可,复杂度 $O(\sqrt{n})$,可以使用 Pollard-Rho 将复杂度降为 $O(\sqrt[4]{n})$。

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inline int getPhi(int p) {
register int ret = p;
for (register int i = 2; i * i <= p; i++) {
(p % i == 0) ? ret = ret / i * (i - 1) : 0;
while (p % i == 0) p /= i;
}
return (p != 1) ? ret / p * (p - 1) : ret;
}

筛 phi

根据性质 1,7 即可线筛 phi。

狄利克雷(Dirichlet)卷积

定义

定义两个数论函数 $f$, $g$ 的 $Dirichlet$ 卷积:
$$(f * g)(n) = \sum_{d | n} f(d) g(\frac{n} {d})$$

一些常见积性函数

  1. 除数函数 $\sigma(n)$ 表示 $n$ 的所有正因子之和,即 $\sigma(n) = \sum_{d | n \text{ and } d > 0} d$,$d(n)$ 表示 $n$ 的正因子个数,即 $d(n) = \sum_{d | n}1$。
  2. 恒等函数 $id(n) = n$。
  3. 单位函数 $\epsilon (n) = \left [ n = 1 \right ]$。
  4. 常函数 $1(n) = 1$。

一些性质

  1. 交换律:$f * g = g * f$
  2. 结合律:$f * g * h = f * (g * h)$
  3. 分配率:$f * (g + h) = f * g + f * h$
  4. 单位元:$f * \epsilon = \epsilon * f$

常见卷积

  1. $d(n) = \sum_{d | n}1 = \sum_{d | n}1(d)1( \frac{n} {d} ) = 1 * 1$
  2. $\sigma (n) = \sum_{d | n}d = \sum_{d | n}d(d)1( \frac{n} {d} ) = d * 1$
  3. $\varphi (n) = \sum_{d | n} \mu (d) \frac{n} {d} = \sum_{d | n} \mu (d)id( \frac{n} {d}) = \mu * id$
  4. $\epsilon (n) = \sum_{d | n} \mu (d) = \sum_{d | n} \mu (d)1( \frac{n} {d}) = \mu * 1$

一些变换

  1. 由于 $\varphi = \mu * id$,$\epsilon = \mu * 1$,所以 $1 * \varphi = 1 * \mu * id$,所以 $1 * \varphi = \epsilon * id = id$,即 $\sum_{d | n} \varphi (d) = n$。
  2. 由于 $\epsilon = \mu * 1$,$\sum_{d | n} \mu (d) = \left [ n = 1 \right ]$

预处理 Dirichlet 卷积

若已知数论函数 $f, g$ ,可以用 $O(n \log n)$ 的时间预处理出 $f * g$。

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int f[MAXN], g[MAXN], h[MAXN];
inline void calculateDirichletProduct(int n) {
for (register int i = 1; i * i <= n; i++) {
for (register int j = i; i * j <= n; j++) {
if (j == i) h[i * j] += f[i] * g[i];
else h[i * j] += f[i] * g[j] + f[j] * g[i];
}
}
}

莫比乌斯函数

定义

$$\mu(n) = \left\{\begin{matrix} 1 & n = 1 \\ (-1)^k & n = p_1p_2 \cdots p_k \\ 0 & otherwise \end{matrix}\right. $$

性质

性质一:莫比乌斯函数是积性函数。
$$\mu(a) \mu(b) = \mu(a \cdot b)$$

性质二:
$$\sum_{d\mid n}\mu(d) = [n = 1]$$
证明
设 $n$ 有 $k (k > 0)$ 个不同的质因子,则 $n$ 所有的质因子中 $\mu \neq 0$ 的只有所有质因子次数都为 $1$ 的因子,质因子个数为 $1$ 的因子有 $\binom{k}{i}$ 个,再利用二项式定理可以得到
$$\sum_{d | n} \mu (d) = \sum_{i = 0}^{k}(-1)^i \cdot \binom{k}{i} = (1 - 1)^k = 0$$
当 $n = 1$ 时原式为 $1$,因此 $\sum_{d | n} \mu (d) = \left [ n = 1 \right ]$

一个结论:$\sum_{d|n}\frac{\mu(d)}{d} = \frac{\varphi(n)}{n}$

莫比乌斯反演

如果 $f(n),\ g(n)$ 是数论函数,且满足:
$$f(n) = \sum_{d\mid n}g(d)$$
则有莫比乌斯反演:
$$g(n) = \sum_{d\mid n}\mu(\frac n d)f(d) = \sum_{d\mid n}\mu(d)f(\frac n d)$$
即 $f = g * 1 \Leftrightarrow g = \mu * f$。
证明:
若 $f = g * 1$,则 $f * \mu = g * (1 * \mu) = g * \epsilon = g$
若 $g = \mu * f$,则 $g * 1 = \mu * f * 1 = \mu * 1 * f = \epsilon * f = f$

变形

$$f(x)=\sum_{x|d}g(d) \Leftrightarrow g(x)=\sum_{x|d}\mu(\frac{d}{x})f(d)$$ $$f(i) = \sum_{d = 1}^{\left\lfloor\frac n i\right\rfloor}g(d\cdot i)\Rightarrow g(i) = \sum_{d = 1}^{\left\lfloor\frac n i\right\rfloor}f(d\cdot i)\mu(d)$$

杜教筛

前置技能

设正整数 $x, y, a, b$,其中 $a, b \leq x, y = \lfloor \frac x a \rfloor$,那么有 $\lfloor \frac y b \rfloor = \lfloor \frac {x} {ab} \rfloor$。

证明:
令 $x = kab + c$,$k, c$ 为非负整数且 $c < ab$,即 $k = \lfloor \frac {x} {ab} \rfloor$。
则 $y = \lfloor \frac x a \rfloor = kb + \lfloor \frac c a \rfloor$,由于 $\lfloor \frac c a \rfloor < b$,所以 $\lfloor \frac y b \rfloor = k$。

设 $f, g$ 为两个数论函数,$t$ 为一个完全积性函数,且 $t(1) = 1$,有
$$f(n) = \sum_{k = 1} ^ {n}t(k)g(\lfloor \frac n k \rfloor) \Leftrightarrow g(n) = \sum_{k = 1} ^ n \mu(k)t(k)f(\lfloor \frac n k \rfloor)$$

证明:
考虑将原式代入,根据结论一得
$$\begin{aligned}\sum_{k = 1} ^ n \mu(k)t(k)f(\lfloor \frac n k \rfloor) &= \sum_{k = 1} ^ n\mu(k)t(k)\sum_{i = 1} ^ {\lfloor \frac n k \rfloor}t(i)g(\lfloor \frac {n} {ki} \rfloor) \\ &= \sum_{j = 1} ^ n g(\lfloor \frac n j \rfloor)\sum_{i | j}\mu(i)t(i)t(\frac j i) \\ &= \sum_{j = 1} ^ n g(\lfloor \frac n j \rfloor)t(j)\sum_{i | j}\mu(i) \end{aligned}$$
由莫比乌斯反演的卷积形式得
$$\begin{aligned}\sum_{k = 1} ^ n\mu(k)t(k)f(\lfloor n k \rfloor) &= \sum_{j = 1} ^ ng(\lfloor \frac n j \rfloor)t(j)\epsilon(j) \\ &= g(n)g(1) \\ &= g(n) \end{aligned}$$
反之亦然。

主要形式

设 $f(n)$ 为一个数论函数,求
$$S(n) = \sum_{i = 1} ^ n f(i)$$

根据函数 $f(n)$ 的性质,构造一个 $S(n)$ 关于 $S(\lfloor \frac n i \rfloor)$ 的递推式,如:
找到一个合适的数论函数 $g(n)$
$$\sum_{i = 1} ^ n \sum_{d | i}f(d)g(\frac i d) = \sum_{T = 1} ^ ng(T)\sum_{i = 1} ^ {\lfloor \frac n T \rfloor}f(i) = \sum_{i = 1} ^ ng(i)S(\lfloor \frac n i \rfloor)$$
可以得到递推式
$$g(1)S(n) = \sum_{i = 1} ^ n(f * g)(i) - \sum_{i = 2} ^ ng(i)S(\lfloor \frac n i \rfloor)$$

「HDU 3579」Hello Kiki

链接

HDU 3579

题解

裸的中国剩余定理,只是不一定互质。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「HDU 3579」Hello Kiki 30-06-2017
* 中国剩余定理
* @author xehoth
*/
#include <bits/stdc++.h>
class InputOutputStream {
private:
static const int BUFFER_SIZE = 1024 * 1024;
char ibuf[BUFFER_SIZE], obuf[BUFFER_SIZE], *s, *t, *oh;
bool isEof;
std::streambuf *cinBuf, *coutBuf;
public:
InputOutputStream(char *in = NULL, char *out = NULL)
: s(ibuf), oh(obuf), isEof(false) {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
in ? freopen(in, "r", stdin) : 0;
out ? freopen(out, "w", stdout) : 0;
cinBuf = std::cin.rdbuf(), coutBuf = std::cout.rdbuf();
t = ibuf + cinBuf->sgetn(ibuf, BUFFER_SIZE);
}
~InputOutputStream() { coutBuf->sputn(obuf, oh - obuf); }
inline char read();
template <typename T>
inline void read(T &);
inline int read(char *);
inline void print(char);
inline void print(const char *);
template <typename T>
inline void print(T);
template <typename T>
inline InputOutputStream &operator>>(T &);
template <typename T>
inline InputOutputStream &operator<<(T);
inline bool hasNext() const { return !isEof; }
} io;
#define long long long
template <typename T>
inline void exgcd(T a, T b, T &g, T &x, T &y) {
!b ? (x = 1, y = 0, g = a) : (exgcd(b, a % b, g, y, x), y -= (a / b) * x);
}
template <typename T>
inline T excrt(T *a, T *m, int n) {
register T M = m[0], ans = a[0], g, x, y;
for (register int i = 1; i < n; i++) {
exgcd(M, m[i], g, x, y);
if ((a[i] - ans) % g) return -1;
x = (a[i] - ans) / g * x % (m[i] / g);
ans = (ans + x * M) % (M = M / g * m[i]);
}
return ans > 0 ? ans : ans + M;
}
const int MAXN = 100005;
int main() {
register int t, n, cas = 1;
for (io >> t; t--; cas++) {
io >> n;
static long a[MAXN], m[MAXN];
for (register int i = 0; i < n; i++) io >> m[i];
for (register int i = 0; i < n; i++) io >> a[i];
io << "Case " << cas << ": " << excrt(a, m, n) << '\n';
}
return 0;
}
inline char InputOutputStream::read() {
s == t ? t = (s = ibuf) + cinBuf->sgetn(ibuf, BUFFER_SIZE) : 0;
return (char)(s == t ? -1 : *s++);
}
template <typename T>
inline void InputOutputStream::read(T &x) {
register char c;
register bool iosig = false;
for (c = read(); !isdigit(c); c = read()) {
if (c == -1) return (void)(isEof = true);
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = (x + (x << 2) << 1) + (c ^ '0');
iosig ? x = -x : 0;
}
inline int InputOutputStream::read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0, isEof = true;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
c == -1 ? isEof = true : 0;
buf[s] = 0;
return s;
}
inline void InputOutputStream::print(char c) {
oh == obuf + BUFFER_SIZE ? (coutBuf->sputn(obuf, BUFFER_SIZE), oh = obuf)
: 0;
*oh++ = c;
}
inline void InputOutputStream::print(const char *s) {
for (; *s; s++) print(*s);
}
template <typename T>
inline void InputOutputStream::print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
template <typename T>
inline InputOutputStream &InputOutputStream::operator>>(T &x) {
read(x);
return *this;
}
template <typename T>
inline InputOutputStream &InputOutputStream::operator<<(T x) {
print(x);
return *this;
}

「BZOJ 1101」Zap

链接

BZOJ 1101

题解

此题就是求
$$\sum_{i = 1} ^ n \sum_{j = 1} ^ m [\gcd(i, j) = k]$$
令 $f(k)$ 为 $\gcd(i, j) = k$ 的个数,$g(k)$ 为 $k \ | \ \gcd(i, j)$ 的对数,则
$$g(k) = \sum_{x = 1} ^ {\lfloor \frac n k \rfloor}f(k \cdot x)$$
若 $i, j$ 能被 $k$ 整除,那么它们可以写成 $i = k * x_1, j = k * x_2$,的形式,我们只需要求有多少对 $x_1, x_2$ 即可,则
$$g(k) = \lfloor \frac n k \rfloor \lfloor \frac m k \rfloor$$
根据莫比乌斯反演的变形可得
$$\begin{aligned}f(k) &= \sum_{x = 1}^{\left\lfloor\frac n k\right\rfloor}\mu(x)g(k\cdot x)\\&=\sum_{x = 1}^{\left\lfloor\frac n k\right\rfloor}\mu(x)\left\lfloor\frac n {kx}\right\rfloor\left\lfloor\frac m {kx}\right\rfloor\end{aligned}$$
由于 $\lfloor \frac n d \rfloor $ 只有 $O(\sqrt{n})$ 种取值,且取值是连续的,所以 $\lfloor \frac {n} {kx} \rfloor, \lfloor \frac {m} {kx} \rfloor$,同时不变的段数有 $O(\sqrt{n} + \sqrt{m})$ 个。
对于相等的段,我们求取 $\mu$ 的前缀和,即可批量计算这个段的答案。

对于位置 $i$,找到下一个相等的位置的代码为 min(n / (n / i), m / (m / i))。对于 $n$,我们要找到最大的 $j$,满足:
$$\lfloor \frac n j \rfloor \geq \lfloor \frac n i \rfloor$$
可以拆掉左边的底:
$$\frac n j \geq \lfloor \frac n i \rfloor$$
化简可得:
$$j \leq \lfloor \frac {n} {\lfloor \frac n i \rfloor} \rfloor$$
所以 $j = \lfloor \frac {n} {\lfloor \frac n i \rfloor} \rfloor$,对于 $m$ 同理,这个复杂度就是 $O(\sqrt{n} + \sqrt{m})$。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 1101」 18-07-2017
* 莫比乌斯反演
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = (x + (x << 2) << 1) + (c ^ '0');
iosig ? x = -x : 0;
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
const int OUT_LEN = 1000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, oh - obuf, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
}
namespace Task {
const int MAXN = 50010;
int prime[MAXN], mu[MAXN], cnt, sum[MAXN];
bool vis[MAXN];
inline void fastLinearSieveMethod(const int n) {
mu[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, mu[i] = -1;
for (register int j = 0, tmp; j < cnt && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
mu[tmp] = 0;
break;
} else {
mu[tmp] = -mu[i];
}
}
}
for (register int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
}
inline int solve(int n, int m) {
if (n > m) std::swap(n, m);
register int ans = 0, pos;
for (register int i = 1; i <= n; i = pos + 1) {
pos = std::min(n / (n / i), m / (m / i));
ans += (sum[pos] - sum[i - 1]) * (n / i) * (m / i);
}
return ans;
}
inline void solve() {
using namespace IO;
register int t;
fastLinearSieveMethod(50000);
read(t);
for (register int i = 0, a, b, c; i < t; i++) {
read(a), read(b), read(c);
print(solve(a / c, b / c)), print('\n');
}
}
}
int main() {
#ifdef DBG
freopen("in.in", "r", stdin);
#endif
Task::solve();
IO::flush();
return 0;
}

「BZOJ 2820」YY的GCD

链接

BZOJ 2820

题解

这道题就是求
$$\sum_{p}\sum_{i = 1} ^ n\sum_{j = 1} ^ m[\gcd(i, j) = p]$$
其中 $p$ 为质数,令
$$f(k) = \sum_{i = 1} ^ n\sum_{j = 1} ^ m[\gcd(i, j) = k]$$
根据上题的推导我们知道:
$$f(k) = \sum_{x = 1}^{\left\lfloor\frac n k\right\rfloor}\mu(x)\left\lfloor\frac n {kx}\right\rfloor\left\lfloor\frac m {kx}\right\rfloor$$
于是
$$\sum_{p}\sum_{i = 1} ^ n\sum_{j = 1} ^ m[\gcd(i, j) = p] = \sum_p \sum_{x = 1}^{\left\lfloor\frac n p\right\rfloor}\mu(x)\left\lfloor\frac n {px}\right\rfloor\left\lfloor\frac m {px}\right\rfloor$$
令 $T = px$,我们在外层枚举 $T$ 然后对每个质因子计算 $\mu$
$$\sum_{T = 1} ^ n\lfloor \frac n T \rfloor \lfloor \frac m T \rfloor\sum_{k | T}\mu(\frac T k)$$

$$f(T) = \sum\limits_{k \mid T} \mu(\frac{T}{k})$$ $$T = p_1 ^ {x_1} \times p_2 ^ {x_2} \times \cdots \times p_n ^ {x_n}$$ $$T' = p_1 ^ {x_1 - 1} \times p_2 ^ {x_2} \times \cdots \times p_n ^ {x_n}$$
考虑线性筛 $\mu$ 的过程,当 $T’ \text{ mod } p_1 = 0$ 时
$$f(T') = \sum_{i = 2} ^ k \mu(\frac {T'} {p_i})$$ $$\begin{aligned}f(T) &= \sum_{i = 1} ^ k \mu(\frac {T} {p_i}) \\ &= \mu(\frac {T} {p_1}) + \sum_{i = 2} ^ k \mu(\frac {T} {p_i}) \\ &= \mu(T') + \sum_{i = 2} ^ k \mu(\frac {T'} {p_i} \times p_1) \\ &= \mu(T') + \sum_{i = 2} ^ k \mu(\frac {T'} {p_i}) \times \mu(p_1) \end{aligned}$$
由于 $\mu(p_1) = -1$,那么
$$f(T) = \mu(T') - f(T')$$

当 $x_1 > 1$ 时,$\mu(p_1 ^ {x_1}) = 0$,则
$$\begin{aligned}f(T) &= \sum_{i = 1} ^ k \mu(\frac {T} {p_i}) \\ &= \mu(\frac {T} {p_1}) + \sum_{i = 2} ^ k \mu(\frac {T} {p_i}) \\ &= \mu(T') + \sum_{i = 2} ^ k \mu(\frac {T} {p_i \times p_1 ^ {x_1}} \times p_1 ^ {x_1}) \\ &= \mu(T') + \sum_{i = 2} ^ k \mu(\frac {T} {p_i \times p_1 ^ {x_1}}) \times \mu(p_1 ^ {x_1}) \\ &= \mu(T') \end{aligned}$$
所以我们线筛预处理后分块回答询问即可。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 2820」 18-07-2017
* 莫比乌斯反演
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = (x + (x << 2) << 1) + (c ^ '0');
iosig ? x = -x : 0;
}
const int OUT_LEN = 1000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
}
namespace Task {
const int MAXN = 10000001;
int prime[MAXN], mu[MAXN], f[MAXN], cnt, sum[MAXN];
bool vis[MAXN];
#define long long long
inline void fastLinearSieveMethod(const int n) {
mu[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, mu[i] = -1, f[i] = 1;
for (register int j = 0, tmp; j < cnt && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
mu[tmp] = 0, f[tmp] = mu[i];
break;
} else {
mu[tmp] = -mu[i], f[tmp] = mu[i] - f[i];
}
}
}
for (register int i = 1; i <= n; i++) sum[i] = sum[i - 1] + f[i];
}
inline long solve(int n, int m) {
n > m ? std::swap(n, m) : (void)0;
register long ans = 0;
for (register int i = 1, pos; i <= n; i = pos + 1) {
pos = std::min(n / (n / i), m / (m / i));
ans += (sum[pos] - sum[i - 1]) * (long)(n / i) * (long)(m / i);
}
return ans;
}
inline void solve() {
using namespace IO;
fastLinearSieveMethod(10000000);
register int t;
read(t);
while (t--) {
register int n, m;
read(n), read(m);
print(solve(n, m)), print('\n');
}
}
#undef long
}
int main() {
Task::solve();
IO::flush();
return 0;
}

「SDOI 2014」数表

链接

BZOJ 3529

题解

令 $\sigma(n)$ 表示 $n$ 的约数和,则第 $i$ 行第 $j$ 列的数为 $\sigma(\gcd(i, j))$,令 $f(k)$ 为 $\gcd(i, j) = k$ 的个数,我们考虑分开枚举每一个 $d = \gcd(i, j)$,它对答案的贡献为 $\sigma(d) \times f(d)$,由 「BZOJ 1101」Zap 的式子我们知道
$$f(k) = \sum_{x = 1}^{\left\lfloor\frac n k\right\rfloor}\mu(x)\left\lfloor\frac n {kx}\right\rfloor\left\lfloor\frac m {kx}\right\rfloor$$
与上题相同,我们令 $T = kx$,我们在外层枚举 $T$ 则
$$\sum_{T = 1} ^ n\lfloor \frac n T \rfloor \lfloor \frac m T \rfloor\sum_{k | T}\mu(\frac T k)$$
故答案
$$\begin{aligned}\sum_{i = 1, i \leq a} ^ n\sigma(i)f(i) &= \sum_{i = 1, i \leq a} ^ n\sigma(i)\sum_{T = 1} ^ n\lfloor \frac n T \rfloor \lfloor \frac m T \rfloor\sum_{i | T}\mu(\frac T i) \\ &= \sum_{T = 1} ^ n\lfloor \frac n T \rfloor \lfloor \frac m T \rfloor \sum_{i | T, i \leq a}\sigma(i)\mu(\frac T i)\end{aligned}$$

由于有 $a$ 的限制,我们不能直接线筛,我们可以把所有的询问按 $a$ 从小到大排序,然后把 $\sigma(x)$ 依次加进去,用树状数组维护前缀和,依次处理询问,时间复杂度为 $O(Q\sqrt{n}\log n + n \log ^ 2n)$。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「SDOI 2014」数表 19-07-2017
* 莫比乌斯反演
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = (x + (x << 2) << 1) + (c ^ '0');
iosig ? x = -x : 0;
}
const int OUT_LEN = 1000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
}
namespace Task {
typedef unsigned int uint;
const int MAXN = 100000;
int prime[MAXN + 10], tot, n, m, max;
bool vis[MAXN + 10];
int t, id[MAXN + 10], sigma[MAXN + 10], mul[MAXN + 10], cnt[MAXN + 10];
uint mu[MAXN + 10], ans[MAXN + 10];
inline void fastLinearSieveMethod(const int n) {
sigma[1] = mu[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) {
prime[tot++] = i, mu[i] = -1, cnt[i] = 1;
mul[i] = i + 1, sigma[i] = i + 1;
}
for (register int j = 0, tmp; j < tot && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
mu[tmp] = 0, cnt[tmp]++, mul[tmp] = mul[i] * prime[j] + 1;
sigma[tmp] = sigma[i] / mul[i] * mul[tmp];
break;
} else {
mu[tmp] = -mu[i], cnt[tmp] = 1, mul[tmp] = prime[j] + 1;
sigma[tmp] = (prime[j] + 1) * sigma[i];
}
}
}
}
struct BinaryIndexedTree {
uint d[MAXN + 10];
inline void modify(int x, uint v) {
for (; x <= max; x += x & -x) d[x] += v;
}
inline uint query(int x) {
register uint ret = 0;
for (; x; x ^= x & -x) ret += d[x];
return ret;
}
} bit;
struct Query {
int l, n, m, id;
inline bool operator<(const Query &b) const { return l < b.l; }
} que[MAXN + 10];
inline bool cmp(const int x, const int y) { return sigma[x] < sigma[y]; }
inline void solve() {
using namespace IO;
read(t);
for (register int i = 1; i <= t; i++) {
read(que[i].n), read(que[i].m), read(que[i].l), que[i].id = i;
max = std::max(max, std::max(que[i].n, que[i].m));
}
fastLinearSieveMethod(max);
for (register int i = 1; i <= max; i++) id[i] = i;
std::sort(id + 1, id + max, cmp);
std::sort(que + 1, que + t + 1);
for (register int i = 1, p = 1; i <= t; i++) {
while (p <= max && sigma[id[p]] <= que[i].l) {
for (register int x = id[p]; x <= max; x += id[p])
bit.modify(x, sigma[id[p]] * mu[x / id[p]]);
p++;
}
n = que[i].n, m = que[i].m, n > m ? std::swap(n, m) : (void)0;
for (register int p = 1, q; p <= n; p = q + 1) {
q = std::min(n / (n / p), m / (m / p));
ans[que[i].id] += (uint)(n / p) * (uint)(m / p) *
(bit.query(q) - bit.query(p - 1));
}
}
for (register int i = 1; i <= t; i++)
print(ans[i] & ((uint)(1 << 31) - 1)), print('\n');
}
}
int main() {
Task::solve();
IO::flush();
return 0;
}

「51 NOD 1244」莫比乌斯函数之和

链接

51 NOD 1244

题解

裸的杜教筛,根据上面所说的杜教筛的主要形式有
$$g(1)S(n) = \sum_{i = 1} ^ n(f * g)(i) - \sum_{i = 2} ^ ng(i)S(\lfloor \frac n i \rfloor)$$
所以我们只需要找到一个合适的数论函数 $g(n)$ 使得我们可以快速计算 $f * g$ 和 $\sum_{i = 2} ^ ng(i)$。
注意到
$$\mu * 1 = \epsilon$$
我们令 $g$ 为常函数 $1$,那么
$$S(n) = \sum_{i = 1} ^ n\epsilon(i) - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) = 1 - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor)$$

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「51 NOD 1244」莫比乌斯函数之和 20-07-2017
* 杜教筛
* @author xehoth
*/
#include <bits/stdc++.h>
namespace Task {
#define long long long
const int MAXN = 4641588 + 100;
int prime[MAXN], mu[MAXN], cnt;
long sum[MAXN];
bool vis[MAXN];
long blockSize, bound, sieveBlockSize;
const int MAX_BLOCK_SIZE = 100000 + 10;
long buc1[MAX_BLOCK_SIZE], buc2[MAX_BLOCK_SIZE];
inline void fastLinearSieveMethod(const int n) {
mu[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, mu[i] = -1;
for (register int j = 0, tmp; j < cnt && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
mu[tmp] = 0;
break;
} else {
mu[tmp] = -mu[i];
}
}
}
for (register int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
}
inline long &get(long x) { return x < blockSize ? buc1[x] : buc2[bound / x]; }
inline long sieveMain(long x) {
if (x <= sieveBlockSize) return sum[x];
register long &cur = get(x);
if (cur != LLONG_MAX) return cur;
register long ret = 1;
for (register long i = 2, pos; i <= x; i = pos + 1)
pos = x / (x / i), ret -= (pos - i + 1) * sieveMain(x / i);
return cur = ret;
}
inline long sieve(long x) {
blockSize = sqrt(x) + 1, bound = x;
std::fill(buc1, buc1 + blockSize + 1, LLONG_MAX);
std::fill(buc2, buc2 + blockSize + 1, LLONG_MAX);
return sieveMain(x);
}
inline void solve() {
std::ios::sync_with_stdio(false), std::cin.tie(NULL), std::cout.tie(NULL);
register long l, r;
std::cin >> l >> r;
sieveBlockSize = pow(r, 2.0 / 3.0) + 1;
fastLinearSieveMethod(sieveBlockSize);
std::cout << sieve(r) - sieve(l - 1);
}
}
int main() {
Task::solve();
return 0;
}

「51 NOD 1239」欧拉函数之和

链接

51 NOD 1239

题解

杜教筛,同上有
$$g(1)S(n) = \sum_{i = 1} ^ n(f * g)(i) - \sum_{i = 2} ^ ng(i)S(\lfloor \frac n i \rfloor)$$
由于
$$\varphi = \mu * id, \epsilon = \mu * 1$$

$$\varphi * 1 = \epsilon * id = id$$
所以令 $g$ 为常函数 $1$,则
$$S(n) = \sum_{i = 1} ^ nid(i) - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) = \frac {n \times (n + 1)} {2} - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor)$$

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「51 NOD 1239」欧拉函数之和 20-07-2017
* 杜教筛
* @author xehoth
*/
#include <bits/stdc++.h>
namespace Task {
#define long long long
const int MAXN = 4641588 + 100;
const int MOD = 1000000007;
int prime[MAXN], phi[MAXN], cnt, sum[MAXN];
bool vis[MAXN];
inline void fastLinearSieveMethod(const int n) {
phi[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, phi[i] = i - 1;
for (register int j = 0, tmp; j < cnt && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
phi[tmp] = phi[i] * prime[j];
break;
} else {
phi[tmp] = phi[i] * (prime[j] - 1);
}
}
}
for (register int i = 1; i <= n; i++) sum[i] = (sum[i - 1] + phi[i]) % MOD;
}
const int MAX_BLOCK_SIZE = 100000 + 10;
long blockSize, sieveBlockSize, bound;
int buc1[MAX_BLOCK_SIZE], buc2[MAX_BLOCK_SIZE];
const int INV_TWO = 500000004;
inline int &get(long x) { return x < blockSize ? buc1[x] : buc2[bound / x]; }
inline int sieveMain(long x) {
if (x <= sieveBlockSize) return sum[x];
register int &cur = get(x);
if (cur != INT_MAX) return cur;
register int ret = ((x % MOD) * ((x + 1) % MOD) % MOD) * INV_TWO % MOD;
for (register long i = 2, pos; i <= x; i = pos + 1)
pos = x / (x / i),
ret = (ret - (pos - i + 1) * (long)sieveMain(x / i) % MOD + MOD) % MOD;
ret = (ret % MOD + MOD) % MOD;
return cur = ret;
}
inline int sieve(long x) {
blockSize = sqrt(x) + 1, bound = x;
std::fill(buc1, buc1 + blockSize + 1, INT_MAX);
std::fill(buc2, buc2 + blockSize + 1, INT_MAX);
return sieveMain(x);
}
inline void solve() {
std::ios::sync_with_stdio(false), std::cin.tie(NULL), std::cout.tie(NULL);
register long n;
std::cin >> n;
sieveBlockSize = pow(n, 2.0 / 3.0) + 1;
fastLinearSieveMethod(sieveBlockSize);
std::cout << sieve(n) << "\n";
}
#undef long
}
int main() {
Task::solve();
return 0;
}

「BZOJ 3944」Sum

链接

BZOJ 3944

题解

把上面两题合在一起就完了…

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 3944」Sum 21-07-2017
* 杜教筛
* @author xehoth
*/
#include <bits/stdc++.h>
namespace Task {
#define long long long
const int MAXN = 2000000 + 100;
int prime[MAXN], mu[MAXN], phi[MAXN], cnt;
long muSum[MAXN], phiSum[MAXN];
bool vis[MAXN];
long blockSize, bound, sieveBlockSize;
const int MAX_BLOCK_SIZE = 46341 + 10;
long buc1[MAX_BLOCK_SIZE], buc2[MAX_BLOCK_SIZE];
inline void fastLinearSieveMethod(const int n) {
mu[1] = 1, phi[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, mu[i] = -1, phi[i] = i - 1;
for (register int j = 0, tmp; j < cnt && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
mu[tmp] = 0, phi[tmp] = phi[i] * prime[j];
break;
} else {
mu[tmp] = -mu[i], phi[tmp] = phi[i] * (prime[j] - 1);
}
}
}
for (register int i = 1; i <= n; i++) muSum[i] = muSum[i - 1] + mu[i];
for (register int i = 1; i <= n; i++) phiSum[i] = phiSum[i - 1] + phi[i];
}
inline long &get(long x) { return x < blockSize ? buc1[x] : buc2[bound / x]; }
inline long sieveMuMain(long x) {
if (x <= sieveBlockSize) return muSum[x];
register long &cur = get(x);
if (cur != LLONG_MAX) return cur;
register long ret = 1;
for (register long i = 2, pos; i <= x; i = pos + 1)
pos = x / (x / i), ret -= (pos - i + 1) * sieveMuMain(x / i);
return cur = ret;
}
inline long sievePhiMain(long x) {
if (x <= sieveBlockSize) return phiSum[x];
register long &cur = get(x);
if (cur != LLONG_MAX) return cur;
register long ret = x * (x + 1) >> 1;
for (register long i = 2, pos; i <= x; i = pos + 1)
pos = x / (x / i), ret -= (pos - i + 1) * sievePhiMain(x / i);
return cur = ret;
}
inline void initSieve(int x) {
blockSize = sqrt(x) + 1, bound = x;
std::fill(buc1, buc1 + blockSize + 1, LLONG_MAX);
std::fill(buc2, buc2 + blockSize + 1, LLONG_MAX);
}
inline void solve() {
std::ios::sync_with_stdio(false), std::cin.tie(NULL), std::cout.tie(NULL);
sieveBlockSize = 2000000 + 1;
fastLinearSieveMethod(sieveBlockSize);
register int T;
std::cin >> T;
while (T--) {
register long n;
std::cin >> n;
initSieve(n);
std::cout << sievePhiMain(n) << ' ';
initSieve(n);
std::cout << sieveMuMain(n) << '\n';
}
}
}
int main() {
Task::solve();
return 0;
}

「HDU 5608」function

链接

HDU 5608

题解

题意就是
$$\sum_{d | n}f(d) = n ^ 2 - 3n + 2$$

$$\sum_{i = 1} ^ nf(i) \text{ mod }10 ^ 9 + 7$$

这题显然是杜教筛,考虑通式
$$g(1)S(n) = \sum_{i = 1} ^ n(f * g)(i) - \sum_{i = 2} ^ ng(i)S(\lfloor \frac n i \rfloor)$$
由于
$$f * 1 = \sum_{d | n}f(d) = n ^ 2 - 3n + 2$$
令 $g$ 为常函数 $1$,则
$$\begin{aligned}S(n) &= \sum_{i = 1} ^ n i ^ 2 - 3i + 2 - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) \\ &= \sum_{i = 1} ^ n i ^ 2 - 3\sum_{i = 1} ^ ni + 2n - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) \\ &= \frac {n * (n + 1) * (2n + 1)} {6} - \frac {3n * (n + 1)} {2} + 2n - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) \\ &= \frac {n * (2n ^ 2 + 3n + 1 - 9n - 9 + 12)} {6} - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) \\ &= \frac {n * (n - 1) * (n - 2)} {3} - \sum_{i = 2} ^ nS(\lfloor \frac n i \rfloor) \end{aligned}$$
接下来我们只需要预处理前 $n ^ {\frac 2 3}$ 的 $S$ 就可以了,令
$$h(n) = \sum_{d | n}f(d) = f * 1$$
由莫比乌斯反演,有
$$f = \mu * h$$

$$\begin{aligned}S(n) &= \sum_{i = 1} ^ nf(i) \\ &= \sum_{i = 1} ^ n\sum_{d | i}h(i)\mu(\frac i d) \\ &= \sum_{i = 1} ^ n\sum_{d | i}(i ^ 2 - 3i + 2)\mu(\frac i d) \\ &= \sum_{i = 1} ^ n\sum_{d | i}(i - 1)(i - 2)\mu(\frac i d) \end{aligned}$$
然后我们就可以 $O(n \log n)$ 预处理前 $n ^ {\frac 2 3}$ 的 $S$ 了,故时间复杂度为 $O(\frac 2 3n ^ {\frac 2 3}\log n + Tn ^ {\frac 2 3})$

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「HDU 5608」function 22-07-2017
* 莫比乌斯反演 + 杜教筛
* @author xehoth
*/
#include <bits/stdc++.h>
namespace Task {
#define long long long
const int MOD = 1e9 + 7;
const int MAXN = 1000000 + 10;
const int INV_THREE = 333333336;
int sieveBlockSize, blockSize, bound;
int buc1[MAXN], buc2[MAXN], f[MAXN], mu[MAXN], cnt, prime[MAXN];
bool vis[MAXN];
inline int &get(int x) { return x < blockSize ? buc1[x] : buc2[bound / x]; }
inline int g(int x) { return (long)(x - 1) * (x - 2) % MOD; }
inline void fastLinearSieveMathod(const int n) {
mu[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, mu[i] = -1;
for (register int j = 0, tmp; j < cnt && (tmp = i * prime[j]) <= n;
j++) {
vis[tmp] = true;
if (i % prime[j] == 0) {
mu[tmp] = 0;
break;
} else {
mu[tmp] = -mu[i];
}
}
}
for (register int i = 1; i <= n; i++)
for (register int j = i; j <= n; j += i)
f[j] = (f[j] + (long)g(i) * mu[j / i] % MOD + MOD) % MOD;
for (register int i = 1; i <= n; i++) f[i] = (f[i] + f[i - 1]) % MOD;
}
inline int sieveMain(int x) {
if (x <= sieveBlockSize) return f[x];
register int &cur = get(x);
if (cur != INT_MAX) return cur;
register int ret =
(long)x * (x - 1) % MOD * (long)(x - 2) % MOD * INV_THREE % MOD;
for (register int i = 2, pos; i <= x; i = pos + 1)
pos = x / (x / i),
ret = (ret - (long)(pos - i + 1) * sieveMain(x / i) % MOD + MOD) % MOD;
return cur = ret;
}
inline int sieve(int x) {
blockSize = sqrt(x) + 1, bound = x;
std::fill(buc1, buc1 + blockSize + 1, INT_MAX);
std::fill(buc2, buc2 + blockSize + 1, INT_MAX);
return sieveMain(x);
}
inline void solve() {
std::ios::sync_with_stdio(false), std::cin.tie(NULL), std::cout.tie(NULL);
sieveBlockSize = 1000000, fastLinearSieveMathod(sieveBlockSize);
register int T, n;
std::cin >> T;
while (T--) std::cin >> n, std::cout << sieve(n) << '\n';
}
#undef long
}
int main() {
Task::solve();
return 0;
}
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