1237 最大公约数之和 V3

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{n}gcd(i,j) \\ =\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^n(gcd(i,j)==d) \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}(gcd(i,j)==1) \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}\sum _{k\mid gcd(i,j)}\mu (k) \\ =\sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}\mu (k)\sum _{i=1}^{\frac{n}{kd}}\sum _{j=1}^{\frac{n}{kd}}1 \\ 套路地设t=kd \\ =\sum _{t=1}^n{\left(\lfloor \frac{n}{t}\rfloor \right)}^2\sum _{d\mid t}d\mu (\frac{t}{d}) \\ =\sum _{t=1}^n{\left(\lfloor \frac{n}{t}\rfloor \right)}^2\varphi (t) \\ 接下来就是杜教筛求\sum _{i=1}^n\varphi (i)了，那这不就是杜教筛水题了嘛。 \end{gathered}$$

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 8e6 + 10, mod = 1000000007;ll phi[N], inv2;int prime[N], cnt;bool st[N];ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}void init() {phi[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;phi[i] = i - 1;}for(int j = 0; j < cnt && 1ll * i * prime[j] < N; j++) {st[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);}}for(int i = 1; i < N; i++) {phi[i] = (phi[i - 1] + phi[i]) % mod;}inv2 = quick_pow(2, mod - 2, mod);
}ll calc(ll x) {x %= mod;return x * (x + 1) % mod * inv2 % mod;
}map<ll, ll> ans_phi;ll get_phi(ll x) {if(x < N) return phi[x];if(ans_phi.count(x)) return ans_phi[x];ll ans = calc(x);for(ll l = 2, r; l <= x; l = r + 1) {r = x / (x / l);ans = (ans - (r - l + 1) % mod * get_phi(x / l) % mod + mod) % mod;}return ans_phi[x] = ans;
}ll calc2(ll x) {x %= mod;return x * x % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll n = read(), ans = 0;init();for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + calc2(n / l) * (get_phi(r) - get_phi(l - 1)) % mod + mod) % mod;}cout << ans << endl;return 0;
}