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A Very Easy Math Problem 题解 ∑a1=1n∑a2=1n⋅⋅⋅∑ax=1n(∏j=1xajk)f(gcd(a1,a2,...,an))gcd(a1,a2,...,an)=∑d=1ndf(d)∑a1=1n∑a2=1n⋅⋅⋅∑ax=1n(∏j=1xajk)[gcd(a1,a2,...,an)=d]=∑d=1ndkx+1f(d)∑a1=1⌊n/d⌋∑a2=1⌊n/d⌋⋅⋅⋅∑ax=1⌊n/d⌋(∏j=1xajk)∑t∣a1,t∣a2,...,t∣anμ(t)=∑d=1ndkx+1f(d)∑

Function 题解 明日更 代码 #include <bits/stdc++.h> using namespace std; typedef long long ll; const int MAX = 1e6 + 10; const int mod = 1e9 + 7; int vis[MAX], prime[MAX], num, mu[MAX], f[MAX]; void makeMobius(int siz) { mu[1] = 1, num = 0; for (int

前置内容 1.a∣ba|ba∣b意为aaa整除bbb 2.[a=b]={1,a=b0,a≠b[a=b]=\begin{cases} 1,a=b\\0,a ≠b \end{cases}[a=b]={1,a=b0,a​=b​ 3.⌊a⌋\lfloor a \rfloor⌊a⌋意为对aaa向下取整 4.当n>k,n>dn>k,n>dn>k,n>d，有⌊⌊n/d⌋k⌋...