CF990G. GCD Counting

题意

有$n$个点的一棵树，每个点有一个权值，对于每一个$i$，问$g(x,y)=i$的$(x,y)$对数$（x\le y)$.$g(x,y)$表示$x$到$y$的路径上的所有$a_i$的$gcd$。

题解

Tip1

点分治：对每一个点统计到各个点的$gcd$，用map存一下。
点分治自己复杂度是$O(nlogn)$,$gcd$、$map$统计，大概有三个$log$
所以总复杂度大约是$(Onlo{g}^{4}n)$

#include
#include
#include
#include
using namespace std;
const int N = 3e5;
struct Edge {
	int u, v, nxt;
}e[N * 2];
int head[N], en, a[N];
void addl(int x, int y) {
	e[++en].u = x, e[en].v = y, e[en].nxt = head[x], head[x] = en; 
}
#define fi first
#define se second
typedef long long ll;
ll ans[N];
bool vis[N];
int n;
int rt;
map<int,int> c, d;
void get(int x, int F, int g) {
	++d[g];++ans[g];
	for(int i = head[x]; i;i = e[i].nxt) {
		int y = e[i].v;
		if(y == F || vis[y]) continue;
		get(y, x, __gcd(g, a[y]));
	}
}

void solve(int x) {
	++ans[a[x]];
	c.clear();
	for(int i = head[x];i; i = e[i].nxt) {
		int y = e[i].v;
		if(vis[y]) continue;
		d.clear();
		get(y, x, __gcd(a[x], a[y]));
		for(auto &p : c)
			for(auto &q : d) {
				int t = __gcd(p.fi, q.fi);
				ans[t] += (ll)p.se * q.se; 
//				printf("%d:++", t);
			}
		for(auto &q : d) c[q.fi] += q.se;
	}
//	for(auto &p: c)
//		cout<<"DD"<
}
int sum;
int siz[N], mx[N];
void getroot(int x, int F) {
	siz[x] = 1;
	mx[x] = 0;	
	for(int i = head[x]; i;i = e[i].nxt) {
		int y = e[i].v;
		if(y == F || vis[y]) continue;
		getroot(y, x);
		siz[x] += siz[y];
		mx[x] = max(mx[x], siz[y]);
	}
	mx[x] = max(mx[x], sum - mx[x]);
	if(mx[x] < mx[rt]) rt = x; 
}
void dfs(int x, int F) {
	vis[x] = 1; solve(x);
	for(int i = head[x]; i; i = e[i].nxt) {
		int y = e[i].v;
		if(vis[y]) continue;
		sum = siz[y];
		rt = 0;
		getroot(y, x);
		dfs(rt, x);
	}
}

map<int,int> t;
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	for(int i = 1; i < n; ++i) {
		int x, y;
		scanf("%d%d",&x,&y);
		addl(x, y);
		addl(y, x);
	}	
	mx[0] = 1e9;
	sum = n;
	getroot(1, 0);
	dfs(rt, 0);
	for(int i = 1; i <= 2e5; ++i)
		if(ans[i]) printf("%d %lld\n", i, ans[i]);
	return 0;
} 

Tip2