洛谷 P1390 公约数的和

题意：

给定$n$,求$\sum_{i=1}^{n}\sum_{j=i+1}^{n}\gcd(i, j)$

思路：

$\sum_{i=1}^{n}\sum_{j=i+1}^{n}\gcd(i, j)$

原式$=$

$\frac{1}{2} * \sum_{i=1}^{n}\sum_{j=1}^{n}\gcd(i, j) - \sum_{i=1}^{n}\gcd(i, i)$

即：

$\frac{1}{2} * \sum_{i=1}^{n}\sum_{j=1}^{n}\gcd(i, j) - \frac{n*(n+1)}{2}$

枚举最大公约数的值：

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

简化式子：

$\frac{1}{2} * \sum_{d=1}^{n} d \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}[\gcd(i, j)=1] - \frac{n*(n+1)}{2}$

将$[\gcd(i, j) = 1]$替换为$\varepsilon(\gcd(i, j))$:

$\frac{1}{2} * \sum_{d=1}^{n} d \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}\varepsilon(\gcd(i, j)) - \frac{n*(n+1)}{2}$

展开$\varepsilon$函数得到：

$\frac{1}{2} * \sum_{d=1}^{n} d \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{k|\gcd(i,j)} \mu(k) - \frac{n*(n+1)}{2}$

变换求和顺序，先枚举$k|\gcd(i, j)$得：

$\frac{1}{2} * \sum_{d=1}^{n} d \sum_{k=1}\mu(k)\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}[k|i]\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}[k|j] - \frac{n*(n+1)}{2}$

已知$1~\lfloor\frac{n}{d}\rfloor$中$k$的倍数有$\lfloor\frac{n}{kd}\rfloor$个：

$\frac{1}{2} * \sum_{d=1}^{n} d \sum_{k=1}\mu(k)\lfloor\frac{n}{kd}\rfloor\lfloor\frac{n}{kd}\rfloor - \frac{n*(n+1)}{2}$

Code：

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
// #include <array>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
// #include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 2e6 + 20;
const int M = 1e5 + 20;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int cnt = 0, primes[N], mu[N], sum[N];
bool st[N];

void get(int n){
    mu[1] = 1;
    for(int i = 2; i <= n; i ++){
        if(!st[i]){
            primes[cnt ++] = i;
            mu[i] = -1;
        }

        for(int j = 0; primes[j] <= n / i; j ++){
            st[i * primes[j]] = true;
            if(i % primes[j] == 0){
                mu[i * primes[j]] = 0;
                break;
            }
            mu[i * primes[j]] = -mu[i];
        }
    }
    sum[0] = 0;
    for(int i = 1; i <= n; i ++){
        sum[i] = sum[i - 1] + mu[i];
    }

}

void solve()
{
    int n;
    sd(n);

    ll ans = 0;
    for(int d = 1; d <= n; d ++){
        ll res = 0;
        int m = n / d;
        for(int l = 1, r; l <= m; l = r + 1){
            r = m / (m / l);
            res = res + (ll)(sum[r] - sum[l - 1]) * (m / l) * (m / l); 
        }
        ans = ans + res * d;
    }
    ans = ans - (ll)n * (n + 1) / 2;
    ans /= 2;
    printf("%lld\n", ans);

}


int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
    // freopen("/home/jungu/code/practice/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    // sd(T);
    get(2000000);
    // init();
    // int cas = 1;
    while (T--)
    {
        // printf("Case #%d:", cas++);
        solve();
    }

    return 0;
}