冷月寒笙的博客

拓展欧几里得模板求解gcd(a,b)=a∗x+b∗y，x与y的值求解gcd(a,b)=a*x+b*y，x与y的值求解gcd(a,b)=a∗x+b∗y，x与y的值由gcd(a,b)=由gcd(a,b)=由gcd(a,b)= gcd(b,agcd(b,agcd(b,a%bbb)))a∗x+b∗y=b∗x+(a−a/b∗b)∗ya*x+b*y=b*x+(a-a/b*b)*ya∗x+b∗y=b∗x+(a−a/b∗b)∗ya∗x+b∗y=a∗y+b∗(x−a/b∗y)a*x+b*y=a*y+b*(x-a/b*

∑i=1di[gcd(i,d)==1]=ϕ(d)∗d2\sum_{i=1}^{d}i[gcd(i,d)==1]=\frac{\phi(d)*d}{2}∑i=1d​i[gcd(i,d)==1]=2ϕ(d)∗d​表示[1,d]中与d互质的数的和

a+b=(a|b)+(a KaTeX parse error: Expected 'EOF', got '&' at position 1: &̲ b)

杜教筛模板记忆化搜索加公式加整数分块g(1)∗s(1)=∑i=1nh(i)+∑d=2ng(d)∗s(⌊nd⌋))g(1)*s(1)=\sum^{n}_{i=1}h(i)+\sum^{n}_{d=2}g(d)*s(\lfloor \frac{n}{d} \rfloor))g(1)∗s(1)=∑i=1n​h(i)+∑d=2n​g(d)∗s(⌊dn​⌋))求ans1=∑i=1nϕ(i)ans1=\sum^{n}_{i=1}\phi(i)ans1=∑i=1n​ϕ(i)即令g(i)=id(i)即可化简可得s(