1220 約數(shù)之和

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{n}d(i,j) \\ =\sum _{i=1}^n\sum _{j=1}^n\sum _{x\mid i}\sum _{y\mid j}(gcd(x,y)=1)\frac{xj}{y} \\ =\sum _{d=1}^{n}d\mu (d)\sum _{i=1}^{\frac{n}{d}}\sum _{x\mid i}\frac{i}{x}\sum _{j=1}^{\frac{n}{d}}\sum _{y\mid j}j \\ =\sum _{d=1}^{n}d\mu (d){?\sum _{i=1}^{\frac{n}{d}}\sum _{x\mid i}x?}^2 \\ =\sum _{d=1}^{n}d\mu (d){?\sum _{x=1}^{n}x\sum _{x\mid i}?}^2 \\ =\sum _{d=1}^{n}d\mu (d){\left(\sum _{x=1}^{n}x\frac{n}{x}\right)}^2 \\ 接下來就是杜教篩求d\mu (d)的前綴和，然后再直接分塊求\sum _{x=1}^{n}x\frac{n}{x}，就可以搞定了。 \end{gathered}$$

代碼

/*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 = 1e6 + 10, mod = 1e9 + 7;int prime[N], cnt;bool st[N];ll mu[N];void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] = (1ll * i * mu[i] + mu[i - 1]) % mod;} }ll calc2(ll l, ll r) {return (l + r) * (r - l + 1) / 2 % mod; }ll calc1(ll n) {ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + calc2(l, r) * (n / l) % mod) % mod;}return ans; }unordered_map<ll, ll> ans_s;ll S(ll n) {if(n < N) return mu[n];if(ans_s.count(n)) return ans_s[n];ll ans = 1;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - calc2(l, r) * S(n / l) % mod + mod) % mod;}return ans_s[n] = (ans + mod) % 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);ll res1 = calc1(n / l), res2 = (S(r) - S(l - 1) + mod) % mod;ans = (ans + res1 * res1 % mod * res2 % mod) % mod;}printf("%lld\n", ans);return 0; }

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