SP5971 LCMSUM

匿名 (未验证) 提交于 2019-12-03 00:10:02

\(\sum_{i=1}^nlcm(i,n)\)

Luogu

SPOJ

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

\[\frac{1}{2}\sum_{i=1}^{n-1}\frac{n^2}{gcd(i,n)}+n\]

\(gcd(i,n)\) 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

\[\frac{n}{2}\sum_{d'|n}\varphi(d')*d'+n\]

\(g(n)=\sum_{d|n}\varphi(d)*d\) ,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; }
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