洛谷 P1390 公约数的和

题目链接

首先我们转化。

\[\begin{aligned} &\sum_{i=1}^{n} \sum_{j=i+1}^{n} \gcd(i,j) \\ &=\frac {\sum_{i=1}^{n}\sum_{j=1}^{n} \gcd(i,j) - \sum_{i=1}^{n} \gcd(i,i)}{2}\\ \end{aligned} \]

然后就是套路

\[\sum_{i=1}^{n} \sum_{j=1}^{n} \gcd(i,j)\\ \sum_{d=1}^{n} d \sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac nd} [\gcd(i,j) = 1]\\ \sum_{d=1}^{n} d \sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac nd} \sum_{p|i,p|j}\mu(p)\\ \sum_{d=1}^{n} d \sum_{p=1}^{\frac nd} \mu(p)\sum_{i=1}^{\frac n{dp}}\sum_{j=1}^{\frac n{dp}} 1\\ \sum_{d=1}^{n} d \sum_{p=1}^{\frac nd} \mu(p)\sum_{i=1}^{\frac n{dp}}\sum_{j=1}^{\frac n{dp}} 1\\ \]

令

\[g(n) = \sum_{p=1}^{n} \mu(p) \sum_{i=1}^{\frac np} \sum_{j=1}^{\frac np} 1\\ g(n) = \sum_{p=1}^{n} \mu(p) \frac np \frac np\\ \]

数论分块求

原式变为

\[\sum_{d=1}^{n} d g(\frac nd)\\ \]

再数论分块。

code

#include<bits/stdc++.h>
using namespace std;
const int N = 2e6;
int prime[N / 10], cnt, p[N + 5], mu[N + 5];
long long g(int n){
    long long ret = 0;
    for(int l = 1 ,r; l <= n; l = r + 1){
        r = min(n, n/(n/l));
        ret = ret + 1ll * (mu[r] - mu[l - 1]) * (n/l) * (n/l);
    }    
    return ret;
}
int n;
int main(){
    p[0] = p[1] = 1; mu[1] = 1;
    for(int i = 2; i <= N; ++ i){
        if(!p[i]) { prime[++ cnt] = i; mu[i] = -1; }
        for(int j = 1; j <= cnt && 1ll * prime[j] * i <= N; ++ j){
            p[prime[j] * i] = 1;
            if(i % prime[j] == 0) break;
            mu[prime[j] * i] = -mu[i];
        }
    }
    for(int i = 1; i <= N; ++ i) mu[i] += mu[i - 1];
    
    scanf("%d",&n);
    long long ans = 0;
    for(int l = 1, r; l <= n; l = r + 1){
        r = min(n, n/(n/l));
        ans = ans + 1ll * (r + l) * (r - l + 1) / 2 * g(n/l);
    }
    ans -= 1ll * (1 + n) * n / 2;
    printf("%lld\n",ans / 2);
    return 0;
}