折戟沉沙

点分治一般按点分为两种情况： 1:过该点,则遍历统计答案 2:不过,递归到子树重心统计 然后合并答案#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fod(i,a,b)

考虑三的倍数的计算: 即t[1]∗t[2]+t[0]2t[1]*t[2]+t[0]^2(每一个点对都计算两次),然后用整个点减去只在子树中出现的即可#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define fo(i,a,b) for(int i=a;i<=b;i