#6229. 這是一道簡單的數學題

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^i\frac{lcm(i,j)}{gcd(i,j)} \\ =(\sum _{i=1}^n\sum _{j=1}^n\frac{lcm(i,j)}{gcd(i,j)}+n)?inv2 \\ 所以重點求\sum _{i=1}^n\sum _{j=1}^n\frac{lcm(i,j)}{gcd(i,j)} \\ =\sum _{i=1}^n\sum _{j=1}^n\frac{ij}{gcd(i,j{)}^2} \\ =\sum _{d=1}^n\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}ij(gcd(i,j)==1) \\ =\sum _{d=1}^n\sum _{k=1}^{\frac{n}{d}}\mu (k)k^2(\sum _{i=1}^{\frac{n}{kd}}i{)}^2 \\ 我們另t=kd，得到 \\ \sum _{t=1}^n(\sum _{i=1}^{\frac{n}{t}}i{)}^2\sum _{k\mid t}\mu (k)k^2 \\ 接下來就是考慮如何在非線性的時間內篩選出\sum _{k\mid t}\mu (k)k^2的前綴和來 \\ 我們設f(n)=(\mu i{d}^2?I)(n)，也就是\sum _{k\mid t}\mu (k)k^2的卷積形式。 \\ g(n)=i{d}^2,顯然有f(n)?g(n)=\mu i{d}^2?I?i{d}^2 \\ \mu i{d}^2?i{d}^2=\sum _{d\mid n}\mu (d)d^2(\frac{n}{d}{)}^2=n^2?=? \\ 所以有f(n)?g(n)=I \\ 套進杜教篩里面去得到S(n)=\sum _{i=1}^{n}I?\sum _{d=2}^n{d}^{2}S(\frac{n}{d}) \end{gathered}$$

代碼

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> #define endl "\n"using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f; const double eps = 1e-7;const int mod = 1e9 + 7, N = 1e6 + 10, inv6 = 166666668, inv2 = 500000004;int prime[N], mu[N], cnt;ll sum[N];bool st[N];ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }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++) {for(int j = i; j < N; j += i) {sum[j] = (sum[j] + 1ll * i * i % mod * mu[i] % mod + mod) % mod;}}for(int i = 1; i < N; i++) {sum[i] = (sum[i] + sum[i - 1]) % mod;} }ll calc1(ll n) {ll ans = 1ll * (1 + n) * n / 2 % mod;return 1ll * ans * ans % mod; }ll calc2(ll n) {return 1ll * n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod; }unordered_map<int, int> ans_s;ll S(int n) {if(n < N) return sum[n];if(ans_s.count(n)) return ans_s[n];ll ans = n;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - 1ll * (calc2(r) - calc2(l - 1) + mod) % mod * S(n / l) % mod + mod) % mod;}return ans_s[n] = ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);// cout << quick_pow(2, mod - 2) << endl;init();ll n, ans = 0;scanf("%lld", &n);for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * calc1(n / l) * ((S(r) - S(l - 1) + mod) % mod) % mod) % mod;}printf("%lld\n", 1ll * (ans + n) * inv2 % mod);return 0; }

總結

以上是生活随笔為你收集整理的#6229. 这是一道简单的数学题（反演 + 杜教筛）的全部內容，希望文章能夠幫你解決所遇到的問題。

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