SP5971 LCMSUM

Posted 2022-11-16 hlw1

题意

求 \(\sum_i=1^nlcm(i,n)\) 。

传送

Luogu

SPOJ

分析

原式可以化为

\[\sum_i=1^n\fraci*ngcd(i,n)\]

由于 \(gcd(i,n)=gcd(n-i,n)\) ，可将原式变形为

\[\frac12(\sum_i=1^n-1\fraci*ngcd(i,n)+\sum_i=n-1^1\fraci*ngcd(i,n))+n\]

两边的 \(sum\) 对应相等，于是有

\[\frac12\sum_i=1^n-1\fracn^2gcd(i,n)+n\]

将 \(gcd(i,n)\) 相等的放在一起统计，枚举 \(gcd(i,n)==d\) ，则 \(gcd(\fracid,\fracnd)==1\) ，故 \(gcd(i,n)==d\) 的数量为 \(\varphi(\fracnd)\) 。

\[\frac12\sum_d|n\frac\varphi(\fracnd)*n^2d+n\]

转换枚举顺序，令 \(d'=\fracnd\) ，上式化为

\[\fracn2\sum_d'|n\varphi(d')*d'+n\]

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。

代码

#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;