three_trees的blog

分析我们设sd(i)sd(i)sd(i)为iii的所有约数之和如果不考虑a对答案的影响：ans=∑i=1n∑j=1msd(gcd(i,j))ans=\sum_{i=1}^{n}\sum_{j=1}^{m}sd(gcd(i,j))ans=i=1∑n​j=1∑m​sd(gcd(i,j))正常的思路，我们枚举gcd(i,j)=dgcd(i,j)=dgcd(i,j)=dans=∑d=1min(n...

分析我们设f(d)f(d)f(d)为gcd⁡=d\gcd=dgcd=d的数量，F(d)F(d)F(d)为gcd⁡\gcdgcd为ddd或者ddd的倍数的数量我们有反演（假设n=3）：f(d)=∑d∣tμ(td)F(t)f(d)=\sum_{d \mid t}\mu(\frac{t}{d})F(t)f(d)=∑d∣t​μ(dt​)F(t)所以这道题难在F(t)F(t)F(t)怎么求：我们设n...

推式子设：num[x]num[x]num[x]表示xxx唯一分解后所有质因子指数之和ans=∑i=1n∑j=1m[num[gcd⁡(i,j)]<=p]\begin{aligned}ans=\sum_{i=1}^{n}\sum_{j=1}^{m}[num[\gcd(i,j)]<=p]\end{aligned}ans=i=1∑n​j=1∑m​[num[gcd(i,j)]&lt...

推式子ans=∑i=1N∑j=1Md(ij)\begin{aligned}ans=\sum_{i=1}^{N}\sum_{j=1}^{M}d(ij)\end{aligned}ans=i=1∑N​j=1∑M​d(ij)​现在有个式子：d(ij)=∑x∣i∑y∣j[gcd⁡(x,y)==1]\begin{aligned}d(ij)=\sum_{x \mid i}\sum_{y \...

题意从[L,H][L,H][L,H]选n个数，且这个n个数的gcd⁡\gcdgcd为k的方案数推式子假设n=3ans=∑x=LH∑y=LH∑z=LH[gcd⁡(x,y,z)==k]\begin{aligned}ans=\sum_{x=L}^{H}\sum_{y=L}^{H}\sum_{z=L}^{H}[\gcd(x,y,z)==k]\end{aligned}ans=x=L∑H​y...

推式子ans=∑i=ab∑j=cd[gcd⁡(i,j)==1]\begin{aligned}ans=\sum_{i=a}^{b}\sum_{j=c}^{d}[\gcd(i,j)==1]\end{aligned}ans=i=a∑b​j=c∑d​[gcd(i,j)==1]​我们给(a,b)(a,b)(a,b)拆成(1,b)−(1,a−1)(1,b)-(1,a-1)(1,b)−(1,a−1...

先贴个代码#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef unsigned long long ull;template <typename T>void out(T x) { cout << x << endl; }const int N...

前置知识数论分块，莫比乌斯函数，和式的相关推导（积累）数论分块这个我写过一篇博客，下面是链接https://blog.youkuaiyun.com/qq_43101466/article/details/100999784莫比乌斯函数μ(n)={1n为1(−1)kk为n的质因子个数0n存在非平方因子{\displaystyle \mu (n)={\begin{cases}1 \quad n为...

推式子ans=∑i=1N∑j=1Mlcm(i,j)=∑i=1N∑j=1Mi∗jgcd⁡(i,j)\begin{aligned}ans = \sum_{i=1}^{N}\sum_{j=1}^{M}lcm(i,j) =\sum_{i=1}^{N}\sum_{j=1}^{M}\frac{i*j}{\gcd(i,j)}\end{aligned}ans=i=1∑N​j=1∑M​lcm(i,j)...

开始推式子ans=∑x=ab∑y=cd[gcd⁡(x,y)=k]\begin{aligned}ans = \sum_{x=a}^{b}\sum_{y=c}^{d}[\gcd(x,y)=k]\end{aligned}ans=x=a∑b​y=c∑d​[gcd(x,y)=k]​还是老套路，我们令f(k)=∑x=ab∑y=cd[gcd⁡(x,y)=k]f(k)=\sum_{x=a}^{b}...

开始推式子ans=∑i=1A∑j=1B∑k=1Cϕ(gcd⁡(i,j2,k3))\begin{aligned}ans = \sum_{i = 1}^{A}\sum_{j = 1}^{B}\sum_{k = 1}^{C}\phi(\gcd(i, j^{2}, k^{3}))\end{aligned} ans=i=1∑A​j=1∑B​k=1∑C​ϕ(gcd(i,j2,k3))​我们令gc...

题意求 ans=∑i=1A∑j=1B[gcd⁡(i,j)==d]ans = \sum_{i = 1}^{A}\sum_{j = 1}^{B}[\gcd(i, j) == d]ans=∑i=1A​∑j=1B​[gcd(i,j)==d]分析我们发现酷似洛谷P2257，题题解链接：https://blog.youkuaiyun.com/qq_43101466/article/details/101...

洛谷-P2257分析本题需要前置知识：数论分块，莫比乌斯反演数论分块，可看https://blog.youkuaiyun.com/qq_43101466/article/details/100999784先介绍一下莫比乌斯函数μ(n)={1n为1(−1)kk为n的质因子个数0n存在非平方因子{\displaystyle \mu (n)={\begin{cases}1 \quad n为1\\(...