TrickGCD Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Problem Description You are given an array A , and Zhu wants to know there are how many d
Problem Description “WTF! While everyone has his girl(gay) friend, I only have my keyboard!” Tired of watching others’ affair, Hillan burst into scream, which made him decide not to hold it back. “A
题目链接:传送门 考虑点 ( x , y ) (x,y) (x,y)对答案的贡献: 设 g c d ( x , y ) = k gcd(x,y)=k gcd(x,y)=k, x = a k , y = b k . x=ak,y=bk. x=ak,y=bk. 若有 x ′ , y ′ x',y' x′,y′满足 ( x ′ , y ′ ) (x',y') (
就是对这个公式的理解 ∑i=1n∑d|iu(d)=1 ∑ i = 1 n ∑ d | i u ( d ) = 1 \sum_{i=1}^n\sum_{d|i}u(d) = 1 首先 ∑d|iu(d)=0,i不等于1的时候 ∑ d | i u ( d ) = 0 , i 不 等 于 1 的 时 候 \sum_{d|i}u(d) = 0,i不等于1的时候
https://www.lydsy.com/JudgeOnline/problem.php?id=2301 这道题阔以作为模板莫比乌斯的模板 并且代码里的函数的意义也与标准的相同: f(d,n,m) f ( d , n , m ) f(d,n,m) 代表小于等于 n、m n 、 m n、m 的 gcd(x,y)=d g c d ( x , y ) = d gcd(x,y)=d 的
第二个样例: N=10,M=10,P=1 N = 10 , M = 10 , P = 1 N=10,M=10,P=1 ans=f(1)+f(2)+f(3)+f(5)+f(7) a n s = f ( 1 ) + f ( 2 ) + f ( 3 ) + f ( 5 ) + f ( 7 ) ans=f(1)+f(2)+f(3)+f(5)+f(7) 每个 f f f 具体的值 以 F
https://vjudge.net/contest/238531#problem/B 题意:给一个 N×N×N N × N × N N\times N\times N的坐标系,从源点 (0,0,0)发出的光线,最多能照到几个坐标点 这道题用的是莫比乌斯反演的倍数那种形式 也就是 F(n)=∑n|df(d) F ( n ) = ∑ n | d f ( d ) F(n)=\s