题目

[CodeForces 439E] Devu and Birthday Celebration

分析

莫比乌斯函数比较重要的性质： μ ∗ 1 = ε \mu * 1 = \varepsilon μ∗1=ε 即 ∑ d ∣ n μ ( d ) = [ n = 1 ] \sum_{d | n}\mu(d) = [n = 1] d∣n∑μ(d)=[n=1]

证明：
设 n = ∏ i = 1 m p i α i n = \prod_{i = 1}^{m} p_i^{\alpha_i} n=∏i=1mpiαi， n ′ = ∏ i = 1 m p i n' = \prod_{i = 1}^mp_i n′=∏i=1mpi，有 ∑ d ∣ n μ ( d ) = ∑ d ∣ n ′ μ ( d ) = ∑ i = 0 m ( m i ) ( − 1 ) i \sum_{d | n}\mu(d) = \sum_{d | n'}\mu(d) = \sum_{i = 0}^{m} \binom{m}{i} (-1)^i d∣n∑μ(d)=d∣n′∑μ(d)=i=0∑m(im)(−1)i 由二项式定理，当 n = 1 n = 1 n=1 时，上式为 1 1 1；当 n ≠ 1 n \neq 1 n=1 时，上式为 [ 1 + ( − 1 ) ] m = 0 [1 + (-1)]^m = 0 [1+(−1)]m=0。

∑ a 1 = 1 n ∑ a 2 = 1 n ⋯ ∑ a f = 1 n [ ∑ i = 1 f a i = n ] [ gcd ⁡ i = 1 f a i = 1 ] = ∑ a 1 = 1 n ∑ a 2 = 1 n ⋯ ∑ a f = 1 n [ ∑ i = 1 f a i = n ] ∑ d ∣ gcd i = 1 f a i μ ( d ) = ∑ d ∣ n μ ( d ) ∑ k 1 = 1 n d ∑ k 2 = 1 n d ⋯ ∑ k f = 1 n d [ ∑ i = 1 f k i = n d ] \begin{aligned} &\sum_{a_1 = 1}^n \sum_{a_2 = 1}^n \cdots \sum_{a_f = 1}^n \left[\sum_{i = 1}^fa_i = n\right] \left[\gcd_{i = 1}^f a_i = 1\right] \\ =& \sum_{a_1 = 1}^n \sum_{a_2 = 1}^n \cdots \sum_{a_f = 1}^n \left[\sum_{i = 1}^f a_i = n\right] \sum_{d |\text{gcd}_{i = 1}^f a_i} \mu(d) \\ =& \sum_{d | n} \mu(d) \sum_{k_1 =1}^{\frac{n}{d}} \sum_{k_2 =1}^{\frac{n}{d}} \cdots \sum_{k_f =1}^{\frac{n}{d}} \left[\sum_{i = 1}^f k_i = \frac{n}{d} \right]\end{aligned} ==a1=1∑na2=1∑n⋯af=1∑n[i=1∑fai=n][i=1gcdfai=1]a1=1∑na2=1∑n⋯af=1∑n[i=1∑fai=n]d∣gcdi=1fai∑μ(d)d∣n∑μ(d)k1=1∑dnk2=1∑dn⋯kf=1∑dn[i=1∑fki=dn]（转换枚举方式， a i = d ⋅ k i a_i = d\cdot k_i ai=d⋅ki，由于 d ∣ a i d | a_i d∣ai，所以 d ∣ ∑ i = 1 f a i d | \sum_{i = 1}^f a_i d∣∑i=1fai 即 d ∣ n d | n d∣n）

预处理莫比乌斯函数，然后 O ( n ) O(\sqrt{n}) O(n ) 枚举因数 + 隔板法即可。

代码

#include <bits/stdc++.h>typedef long long LL;template <const int _MOD> struct ModNumber {int x;inline ModNumber() { x = 0; }inline ModNumber(const int &y) { x = y; }inline int Int() { return x; }inline ModNumber Pow(int y) const { register int ret = 1, tmp = x; while (y) { if (y & 1) ret = ((LL)ret * tmp) % _MOD; y >>= 1; tmp = ((LL)tmp * tmp) % _MOD; } return ModNumber(ret); }inline bool operator == (const ModNumber &y) const { return x == y.x; }inline bool operator != (const ModNumber &y) const { return x != y.x; }inline bool operator < (const ModNumber &y) const { return x < y.x; }inline bool operator > (const ModNumber &y) const { return x > y.x; }inline bool operator <= (const ModNumber &y) const { return x <= y.x; }inline bool operator >= (const ModNumber &y) const { return x >= y.x; }inline ModNumber operator - () const { return _MOD - x; }inline ModNumber operator + (const ModNumber &y) const { return (x + y.x >= _MOD) ? (x + y.x - _MOD) : (x + y.x); }inline ModNumber operator - (const ModNumber &y) const { return (x - y.x < 0) ? (x - y.x + _MOD) : (x - y.x); }inline ModNumber operator * (const ModNumber &y) const { return ModNumber((LL)x * y.x % _MOD); }inline ModNumber operator / (const ModNumber &y) const { return *this * y.Pow(_MOD - 2); }inline ModNumber operator ^ (const int &y) const { return Pow(y); }inline void operator += (const ModNumber &y) { *this = *this + y; }inline void operator *= (const ModNumber &y) { *this = *this * y; }inline void operator -= (const ModNumber &y) { *this = *this - y; }inline void operator /= (const ModNumber &y) { *this = *this / y; }inline void operator ^= (const int &y) const { *this = *this ^ y; }inline bool operator == (const int &y) const { return x == y; }inline bool operator != (const int &y) const { return x != y; }inline bool operator < (const int &y) const { return x < y; }inline bool operator > (const int &y) const { return x > y; }inline bool operator <= (const int &y) const { return x <= y; }inline bool operator >= (const int &y) const { return x >= y; }inline ModNumber operator + (const int &y) const { return (x + y >= _MOD) ? (x + y - _MOD) : (x + y); }inline ModNumber operator - (const int &y) const { return (x - y < 0) ? (x - y + _MOD) : (x - y); }inline ModNumber operator * (const int &y) const { return ModNumber((LL)x * y % _MOD); }inline ModNumber operator / (const int &y) const { return *this * ModNumber(y).Pow(_MOD - 2); }inline void operator += (const int &y) { *this = *this + y; }inline void operator *= (const int &y) { *this = *this * y; }inline void operator -= (const int &y) { *this = *this - y; }inline void operator /= (const int &y) { *this = *this / y; }
};const int MAXN = 2 * 100000;
const int MOD = 1000000007;typedef ModNumber<MOD> Int;Int Fac[MAXN + 5], Inv[MAXN + 5];Int Mu[MAXN + 5];
bool Vis[MAXN + 5];
std::vector<int> Primes;void Init(int n) {Mu[1] = 1;for (int i = 2; i <= n; i++) {if (!Vis[i])Mu[i] = MOD - 1, Primes.push_back(i);for (int j = 0; j < (int)Primes.size() && i * Primes[j] <= n; j++) {Vis[i * Primes[j]] = true;if (i % Primes[j] == 0) {Mu[i * Primes[j]] = 0;break;}Mu[i * Primes[j]] = -Mu[i];}}Fac[0] = 1;for (int i = 1; i <= n; i++)Fac[i] = Fac[i - 1] * i;Inv[n] = Fac[n] ^ (MOD - 2);for (int i = n - 1; i >= 0; i--)Inv[i] = Inv[i + 1] * (i + 1);
}Int C(int n, int m) {if (n < m) return 0;return Fac[n] * Inv[m] * Inv[n - m];
}int main() {Init(MAXN);int Q; scanf("%d", &Q);while (Q--) {Int Ans = 0;int N, F; scanf("%d%d", &N, &F);for (int i = 1; i * i <= N; i++) {if (N % i) continue;int j = N / i; Ans += Mu[i] * C(j - 1, F - 1);if (j != i) Ans += Mu[j] * C(i - 1, F - 1);}printf("%d\n", Ans.x);}return 0;
}

[CodeForces 439E] Devu and Birthday Celebration（莫比乌斯反演） | 错题本相关推荐

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[CodeForces 439E] Devu and Birthday Celebration（莫比乌斯反演） | 错题本

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