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

题目

[CodeForces 439E] Devu and Birthday Celebration

分析

莫比乌斯函数比较重要的性质：$$\mu \ast 1=\epsilon$$ 即 $$\sum _{d|n}\mu (d)=[n=1]$$

证明：
设 $n={\prod }_{i=1}^m{p}_i^{{\alpha }_i}$，$n^{\prime }={\prod }_{i=1}^m{p}_i$，有 $$\sum _{d|n}\mu (d)=\sum _{d|n^{\prime }}\mu (d)=\sum _{i=0}^m\left(\genfrac{}{}{0ex}{}{m}{i}\right)(-1{)}^i$$ 由二项式定理，当 $n=1$ 时，上式为 $1$；当 $n\ne 1$ 时，上式为 $[1+(-1){]}^m=0$。

$$\begin{array}{rl}& \sum _{a_1=1}^n\sum _{a_2=1}^n\cdots \sum _{a_f=1}^n\left[\sum _{i=1}^f{a}_i=n\right]\left[\underset{i=1}{\overset{f}{\gcd }}{a}_i=1\right]\\ =& \sum _{a_1=1}^n\sum _{a_2=1}^n\cdots \sum _{a_f=1}^n\left[\sum _{i=1}^f{a}_i=n\right]\sum _{d|{\text{gcd}}_{i=1}^f{a}_i}\mu (d)\\ =& \sum _{d|n}\mu (d)\sum _{k_1=1}^{\frac{n}{d}}\sum _{k_2=1}^{\frac{n}{d}}\cdots \sum _{k_f=1}^{\frac{n}{d}}\left[\sum _{i=1}^f{k}_i=\frac{n}{d}\right]\end{array}$$（转换枚举方式，$a_i=d\cdot k_i$，由于 $d|a_i$，所以 $d|{\sum }_{i=1}^f{a}_i$ 即 $d|n$）

预处理莫比乌斯函数，然后 $O(\sqrt{n})$ 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;
}