GCD of Sequence

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题目链接

题意

给一个序列，让你计算有多少序列$b$满足:
1.每个位置。
2.。
3.和原来数列恰好有$k$个位置数字不同。

思路

思路不难想，计算出$F(d)$，然后去暴力计算每一个$f(d)$。

其中$c$为数字$d$的倍数出现的次数。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50

const int N = 3e5 + 5;
int a[N];
int prime[N];
int tot, vis[N];
int miu[N];
ll F[N];
void pre(int n)
{
	miu[1] = 1;
	tot = 0;
	for (int i = 2; i <= n; i++)
	{
		if (!vis[i])
		{
			prime[++tot] = i;
			miu[i] = -1;
		}
		for (int j = 1; j <= tot && i * prime[j] <= n; j++)
		{
			vis[i * prime[j]] = 1;
			if (i % prime[j] == 0)
			{
				miu[i * prime[j]] = 0;
				break;
			}
			miu[i * prime[j]] = -miu[i];
		}
	}
}
int cnt[N];
const ll mod = 1e9 + 7;
ll Pow(ll a, ll b)
{
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = ans * a % mod;
		a = a * a % mod;
		 b >>= 1;
	}
	return ans;
}
ll fac[N];
ll C(int n, int m)
{
	return fac[n] * Pow(fac[n - m] * fac[m] % mod, mod - 2) % mod;
}

int main()
{
	fac[0] = 1;
	for (int i = 1; i <= 300000; i++)
		fac[i] = fac[i - 1] * 1ll * i % mod;
	pre(300000);
	int n, m, k;
	while (~scanf("%d%d%d", &n, &m, &k))
	{
		memset(cnt, 0, sizeof(cnt));
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
			cnt[a[i]]++;
		}
		for (int i = 1; i <= m; i++)
		{
			ll c = 0;
			for (ll d = i; d <= m; d += i)
			{
				c += cnt[d];
			}
			if (n - c > k)
				F[i] = 0;
			else
				F[i] = C(c, k - n + c) * Pow(m / i - 1ll, k - n + c) % mod * Pow(m / i, n - c) % mod;
			//dbg(i, F[i], k, n, c);
		}
		for (int i = 1; i <= m; i++)
		{
			ll ans = 0;
			for (ll d = i; d <= m; d += i)
				ans = (ans + miu[d / i] * F[d] % mod + mod) % mod;
			printf("%lld%c", ans, i == m? '\n': ' ');
		}
	}
	return 0;
}

这里提供另一种思路计算某个数字的倍数的出现次数的方法，用莫比乌斯函数容斥一下。
但让我百思不得其解的是这种方法竟然比上面要慢。
如果有人可以帮我证明这两种方法的优劣性，欢迎联系我。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50

const int N = 3e5 + 5;
int a[N];
int prime[N];
int tot, vis[N];
int miu[N];
ll F[N];
void pre(int n)
{
	miu[1] = 1;
	tot = 0;
	for (int i = 2; i <= n; i++)
	{
		if (!vis[i])
		{
			prime[++tot] = i;
			miu[i] = -1;
		}
		for (int j = 1; j <= tot && i * prime[j] <= n; j++)
		{
			vis[i * prime[j]] = 1;
			if (i % prime[j] == 0)
			{
				miu[i * prime[j]] = 0;
				break;
			}
			miu[i * prime[j]] = -miu[i];
		}
	}
}
int cnt[N];
const ll mod = 1e9 + 7;
ll Pow(ll a, ll b)
{
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = ans * a % mod;
		a = a * a % mod;
		 b >>= 1;
	}
	return ans;
}
ll fac[N];
ll dfs(int p, int m)
{
	if (vis[p])
		return cnt[p];
	for (int i = 2; p * i <= m; i++)
		if (miu[i])
			cnt[p] = (cnt[p] - miu[i] * dfs(p * i, m) % mod + mod) % mod;
	vis[p] = 1;
	//dbg(p, cnt[p]);
	return cnt[p];
}
ll C(int n, int m)
{
	return fac[n] * Pow(fac[n - m] * fac[m] % mod, mod - 2) % mod;
}

int main()
{
	fac[0] = 1;
	for (int i = 1; i <= 300000; i++)
		fac[i] = fac[i - 1] * 1ll * i % mod;
	pre(300000);
	int n, m, k;
	while (~scanf("%d%d%d", &n, &m, &k))
	{
		memset(cnt, 0, sizeof(cnt));
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
			cnt[a[i]]++;
		}
		for (int i = 1; i <= m; i++)
			vis[i] = 0;
		dfs(1, m);
		for (int i = 1; i <= m; i++)
		{
			/*
			ll c = 0;
			for (ll d = i; d <= m; d += i)
			{
				c += cnt[d];
			}
			*/
			ll c = cnt[i];
			if (n - c > k)
				F[i] = 0;
			else
				F[i] = C(c, k - n + c) * Pow(m / i - 1ll, k - n + c) % mod * Pow(m / i, n - c) % mod;
			//dbg(i, F[i], k, n, c);
		}
		for (int i = 1; i <= m; i++)
		{
			ll ans = 0;
			for (ll d = i; d <= m; d += i)
				ans = (ans + miu[d / i] * F[d] % mod + mod) % mod;
			printf("%lld%c", ans, i == m? '\n': ' ');
		}
	}
	return 0;
}

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