莫比乌斯反演

概念可借鉴参考博客，具体证明见B站电子科大数学学院的视频（貌似用到狄利克雷卷积，反正我没看懂）
前置知识点：整数分块
貌似除了算多少gcd(i,j)=n外，还有别的用处。

矩阵求和

原题地址

代数学得不好（毕竟不是数院的），大概写了写推导过程

#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
#define maxn 10000010
#define mkp make_pair
#define inf 1e6
typedef long long ll;
const ll mod = 1e9 + 7;
const double pi = acos(-1.0);
ll sum[maxn], pri[maxn], qq[maxn];
ll n;
int tot = 0, vis[maxn], mu[maxn];
void init()
{
    ll i, j;
    mu[1] = 1;
    for (i = 2; i <= n; i++)
    {
        if (!vis[i])
        {
            pri[++tot] = i;
            mu[i] = -1;
        }
        for (j = 1; j <= tot && pri[j] * i <= n; j++)
        {
            vis[i * pri[j]] = 1;
            if (i % pri[j]) mu[i * pri[j]] = -mu[i];
            else
            {
                mu[i * pri[j]] = 0;
                break;
            }
        }
    }
    for (i = 1; i <= n; i++) sum[i] = (sum[i - 1] + mu[i] + mod) % mod;
    for (i = 1; i <= n; i++)
    {
        qq[i] = (qq[i - 1] + i * i % mod) % mod;
    }
}

ll solve(ll tx)
{
    ll l, r, ans = 0;
    for (l = 1; l <= tx; l = r + 1)
    {
        ll tw = tx / l;
        r = tx / tw;
        ans = (ans + (sum[r] - sum[l - 1] + mod) % mod * (tw * tw % mod) % mod) % mod;
    }
    return ans;
}

int main() 
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll i, j, r, l, ans = 0;
    cin >> n;
    init();
    for (l = 1; l <= n; l = r + 1)
    {
        ll tx = n / l;
        r = n / tx;
        ans = (ans + (qq[r] - qq[l - 1] + mod) % mod * solve(tx) % mod) % mod;
    }
    cout << ans << endl;
    return 0;
}