Lynstery's blog

题意求∑Ri=Lμ(i)\sum_{i=L}^R\mu(i)，L,R≤1011L,R\le10^{11}题解杜教筛模板题。 g(1)S(n)=∑i=1n(f∗g)(i)−∑i=2ng(i)S(⌊ni⌋)g(1)S(n)=\sum_{i=1}^n(f*g)(i)-\sum_{i=2}^ng(i)S(\lfloor\frac{n}{i}\rfloor) μ∗1=ϵ\mu∗1=\epsilon，gg取

题意求∑ni=1ϕ(i)\sum_{i=1}^n\phi(i)题解模板题 S(n)=(1+n)∗n2−∑i=2nS(⌊ni⌋)S(n)=\frac{(1+n)*n}{2}-\sum_{i=2}^nS(\lfloor\frac{n}{i}\rfloor)#include<cstdio>#include<map>#include<cstring>#include<algorithm>#incl

题意给定n, 求∑ni=1ϕ(i)\sum_{i=1}^n \phi(i)和∑ni=1μ(i)\sum_{i=1}^n \mu(i) n≤231−1n \le 2^{31-1}题解假装做了一道新题 这道题=这题+这题水啊水#include<cstdio>#include<map>#include<cstring>#include<algorithm>#include <tr1/unord

题意题解我们设 g(i)=i2−3i+2g(i)=i^2-3i+2，题目告诉我们的就是 g=f∗1g=f*1。然后就可以根据杜教筛的思路搞了： ∑i=1ng(i)=∑i=1n∑d|if(d)=∑i=1n∑d⌊ni⌋f(d)=∑i=1nS(⌊ni⌋)\sum_{i=1}^ng(i)=\sum_{i=1}^n \sum_{d|i}f(d)=\sum_{i=1}^n\sum_d^{\lfloor\fra

题意求∑ni∑njgcd(i,j)\sum_i^n\sum_j^ngcd(i,j) n≤1010n\le10^{10}题解∑in∑jngcd(i,j)=∑d=1n∑i=1n∑j=1n[gcd(i,j)=d]∗d=∑d=1n∑i=1⌊nd⌋∑j=1⌊nd⌋[gcd(i,j)=1]∗d\sum_i^n\sum_j^ngcd(i,j)=\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}

题意求∑ni=1∑nj=1lcm(i,j)\sum_{i=1}^n \sum_{j=1}^n lcm(i,j) n≤1010n \le 10^{10}题解自己推到后面就不会了，然后去找题解…… 大概就是 ∑i=1n∑j=1nijgcd(i,j)=∑d=1n∑i=1⌊nd⌋∑j=1⌊nd⌋[gcd(i,j)=1]i∗j∗d2d\sum_{i=1}^n \sum_{j=1}^n \frac{ij}