求 \(\sum_{i=1}^nlcm(i,n)\) 。
原式可以化为
\[\sum_{i=1}^n\frac{i*n}{gcd(i,n)}\]
由于 \(gcd(i,n)=gcd(n-i,n)\) ,可将原式变形为
\[\frac{1}{2}(\sum_{i=1}^{n-1}\frac{i*n}{gcd(i,n)}+\sum_{i=n-1}^{1}\frac{i*n}{gcd(i,n)})+n\]
两边的 \(sum\) 对应相等,于是有
\[\frac{1}{2}\sum_{i=1}^{n-1}\frac{n^2}{gcd(i,n)}+n\]
将 \(gcd(i,n)\) 相等的放在一起统计,枚举 \(gcd(i,n)==d\) ,则 \(gcd(\frac{i}{d},\frac{n}{d})==1\) ,故 \(gcd(i,n)==d\) 的数量为 \(\varphi(\frac{n}{d})\) 。
\[\frac{1}{2}\sum_{d|n}\frac{\varphi(\frac{n}{d})*n^2}{d}+n\]
转换枚举顺序,令 \(d'=\frac{n}{d}\) ,上式化为
\[\frac{n}{2}\sum_{d'|n}\varphi(d')*d'+n\]
设 \(g(n)=\sum_{d|n}\varphi(d)*d\) ,已知 \(g(n)\) 为积性函数,则可以预处理出答案,直接输出即可。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define N 1000000 #define il inline #define re register #define tie0 cin.tie(0),cout.tie(0) #define fastio ios::sync_with_stdio(false) #define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout) using namespace std; typedef long long ll; template <typename T> inline void read(T &x) { T f = 1; x = 0; char c; for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1; for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48); x *= f; } int n; ll ans[N+5]; int phi[N+5], prime[N+5]; bool vis[N+5]; void get_phi() { int cnt = 0; phi[1] = 1; for (int i = 2; i <= N; ++i) { if (!vis[i]) prime[++cnt] = i, phi[i] = i - 1; for (int j = 1; j <= cnt && i * prime[j] <= N; ++j) { vis[i*prime[j]] = 1; if (i % prime[j] == 0) { phi[i*prime[j]] = phi[i] * prime[j]; break; } phi[i*prime[j]] = phi[i] * (prime[j] - 1); } } } void pre() { for (int i = 1; i <= N; ++i) for (int j = 1; j * i <= N; ++j) ans[i*j] += 1ll * j * phi[j] / 2; for (int i = 1; i <= N; ++i) ans[i] = 1ll * ans[i] * i + i; } int main() { get_phi(); pre(); int t; read(t); while (t--) { read(n); printf("%lld\n", ans[n]); } return 0; }
来源:博客园
作者:siye1989
链接:https://www.cnblogs.com/hlw1/p/11562195.html