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  • CF809E Surprise me!

    题面

    题解

    这道题目的话,推式子比较休闲,写起来。。。

    首先上套路,根据(varphi)的一些性质,我们可以证明(varphi(ij) = frac{varphi(i)varphi(j)gcd(i, j)}{varphi(gcd(i,j))})

    开推:首先设( extbf{f}(x) = frac x {varphi(x)}, extbf g = extbf f * mu, p_{a_i} = i)

    那么有(我的这种推法是学的rqy的比较简单的,建议参阅):

    [egin{aligned} &sum_{i=1}^nsum_{j=1}^nvarphi(a_i*a_j)cdot mathrm{dist}(i,j) \ =& sum_{i=1}^nsum_{j=1}^nvarphi(i)varphi(j)mathrm{dist}(p_i, p_j) extbf f(gcd(i,j)) \ =& sum_{i=1}^nsum_{j=1}^nvarphi(i)varphi(j)mathrm{dist}(p_i, p_j)sum_{dmid i, dmid j} extbf g(d) \ =& sum_{d=1}^n extbf g(d)sum_{dmid i}sum_{dmid j}varphi(i)varphi(j)mathrm{dist}(p_{i},p_{j}) end{aligned} ]

    ( extbf g)可以枚举倍数求,复杂度调和级数。

    然后后面那一坨可以枚举倍数,将所有是(d)的倍数的点全部拉出来建一棵虚树,做一遍树形(mathrm{dp})就可以了。

    虚树的总点数在(mathrm{O}(nlog n))级别,所以总复杂度约为(mathrm{O}(nlog^2n))

    代码

    #include<cstdio>
    #include<cstring>
    #include<cctype>
    #include<algorithm>
    #define RG register
    #define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
    #define clear(x, y) memset(x, y, sizeof(x))
    
    inline int read()
    {
    	int data = 0, w = 1; char ch = getchar();
    	while(ch != '-' && (!isdigit(ch))) ch = getchar();
    	if(ch == '-') w = -1, ch = getchar();
    	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
    	return data * w;
    }
    
    const int maxn(2e5 + 10), Mod(1e9 + 7);
    int fastpow(int x, int y)
    {
    	int ans = 1;
    	for(; y; y >>= 1, x = 1ll * x * x % Mod)
    		if(y & 1) ans = 1ll * ans * x % Mod;
    	return ans;
    }
    
    namespace Numbers
    {
    	int prime[maxn], cnt, mu[maxn], phi[maxn], not_prime[maxn], iphi[maxn];
    	void Init(int n)
    	{
    		mu[1] = phi[1] = not_prime[1] = 1;
    		for(RG int i = 2; i <= n; i++)
    		{
    			if(!not_prime[i]) prime[++cnt] = i, phi[i] = i - 1, mu[i] = Mod - 1;
    			for(RG int j = 1; j <= cnt && i * prime[j] <= n; j++)
    			{
    				not_prime[i * prime[j]] = 1;
    				if(i % prime[j]) mu[i * prime[j]] = Mod - mu[i],
    					phi[i * prime[j]] = phi[i] * phi[prime[j]];
    				else { phi[i * prime[j]] = phi[i] * prime[j]; break; }
    			}
    		}
    		for(RG int i = 1; i <= n; i++)
    			iphi[i] = fastpow(phi[i], Mod - 2);
    	}
    }
    
    namespace Vtree
    {
    	struct edge { int next, to, dis; } e[maxn << 1];
    	int head[maxn], e_num, fa[maxn], size[maxn], heavy[maxn], stk[maxn << 1];
    	int pos[maxn], end_pos[maxn], belong[maxn], cnt, p[maxn << 1], K, dep[maxn];
    	inline void add_edge(int from, int to, int dis = 0)
    	{
    		e[++e_num] = (edge) {head[from], to, dis};
    		head[from] = e_num;
    	}
    
    	void dfs(int x)
    	{
    		size[x] = 1;
    		for(RG int i = head[x]; i; i = e[i].next)
    		{
    			int to = e[i].to; if(to == fa[x]) continue;
    			fa[to] = x, dep[to] = dep[x] + 1; dfs(to); size[x] += size[to];
    			if(size[heavy[x]] < size[to]) heavy[x] = to;
    		}
    	}
    
    	void dfs(int x, int chain)
    	{
    		belong[x] = chain, pos[x] = ++cnt;
    		if(heavy[x]) dfs(heavy[x], chain);
    		for(RG int i = head[x]; i; i = e[i].next)
    		{
    			int to = e[i].to;
    			if(to == fa[x] || to == heavy[x]) continue;
    			dfs(to, to);
    		}
    		end_pos[x] = cnt;
    	}
    
    	int LCA(int x, int y)
    	{
    		for(; belong[x] != belong[y]; x = fa[belong[x]])
    			if(pos[belong[x]] < pos[belong[y]]) std::swap(x, y);
    		return pos[x] < pos[y] ? x : y;
    	}
    
    	int Dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[LCA(x, y)]; }
    	inline bool cmp(int x, int y) { return pos[x] < pos[y]; }
    	void build()
    	{
    		e_num = 0; std::sort(p + 1, p + K + 1, cmp);
    		for(RG int i = K; i > 1; i--) p[++K] = LCA(p[i], p[i - 1]);
    		p[++K] = 1; std::sort(p + 1, p + K + 1, cmp);
    		K = std::unique(p + 1, p + K + 1) - p - 1;
    		for(RG int i = 1, top = 0; i <= K; i++)
    		{
    			while(top && end_pos[stk[top]] < pos[p[i]]) --top;
    			add_edge(stk[top], p[i], Dis(stk[top], p[i])); stk[++top] = p[i];
    		}
    	}
    }
    
    int f[maxn], g[maxn], s[maxn], a[maxn], b[maxn], c[maxn];
    void dfs(int x)
    {
    	f[x] = g[x] = 0, s[x] = b[x];
    	for(RG int i = Vtree::head[x]; i; i = Vtree::e[i].next)
    	{
    		int to = Vtree::e[i].to, dis = Vtree::e[i].dis; dfs(to);
    		g[x] = (((g[x] + g[to]) % Mod + 1ll * s[to] * (f[x] +
    			1ll * s[x] * dis % Mod) % Mod) % Mod
    				+ 1ll * f[to] * s[x] % Mod) % Mod;
    		s[x] = (s[x] + s[to]) % Mod;
    		f[x] = ((f[x] + f[to]) % Mod + 1ll * s[to] * dis % Mod) % Mod;
    	}
    	Vtree::head[x] = 0, b[x] = 0;
    }
    
    int n;
    int main()
    {
    	n = read();
    	for(RG int i = 1; i <= n; i++) a[read()] = i;
    	for(RG int i = 1, x, y; i < n; i++)
    		x = read(), y = read(), Vtree::add_edge(x, y), Vtree::add_edge(y, x);
    	Vtree::dfs(1); Vtree::dfs(1, 1); memset(Vtree::head, 0, sizeof Vtree::head);
    	Numbers::Init(n); using Numbers::mu; using Numbers::iphi;
    	for(RG int d = 1; d <= n; d++)
    		for(RG int i = d; i <= n; i += d)
    			c[i] = (c[i] + 1ll * d * mu[i / d] % Mod * iphi[d] % Mod) % Mod;
    	int ans = 0;
    	for(RG int i = 1; i <= n; i++)
    	{
    		Vtree::K = 0;
    		for(RG int j = i; j <= n; j += i)
    			Vtree::p[++Vtree::K] = a[j],
    			b[a[j]] = Numbers::phi[j];
    		Vtree::build(); dfs(1);
    		ans = (ans + 2ll * c[i] * g[1] % Mod) % Mod;
    	}
    	ans = 1ll * ans * fastpow(n, Mod - 2) % Mod * fastpow(n - 1, Mod - 2) % Mod;
    	printf("%d
    ", ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cj-xxz/p/10597132.html
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