1227 平均最小公倍數

推式子

$$\begin{gathered} S(n)=\sum _{i=1}^n\sum _{j=1}^i\frac{lcm(i,j)}{i} \\ =\sum _{i=1}^n\sum _{j=1}^i\frac{ij}{igcd(i,j)} \\ =\sum _{i=1}^n\sum _{j=1}^i\frac{j}{gcd(i,j)} \\ =\sum _{i=1}^n\sum _{d=1}^i\sum _{j=1}^i\frac{j}{d}(gcd(i,j)==d) \\ =\sum _{i=1}^n\sum _{d=1}^i\sum _{j=1}^{\frac{i}{d}}j(gcd(j,\frac{i}{d})==1) \\ =\sum _{i=1}^n\sum _{d=1}^i\frac{\frac{i}{d}?(\frac{i}{d})+(\frac{i}{d}==1)}{2} \\ =\sum _{d=1}^n\sum _{i=1}^{\frac{n}{d}}\frac{i?(i)+(i==1)}{2} \\ 接下來就是杜教篩求\sum _{i=1}^{n}i?(i)了，g(1)S(n)=\sum _{i=1}^n(f?g)(i)?\sum _{i=2}^{n}f(i)S(\frac{n}{i}) \\ 另f(i)=i，即可求得S(n)=\sum _{i=1}^n{i}^2?\sum _{i=2}^{n}iS(\frac{n}{d}) \end{gathered}$$

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;ll phi[N], inv2, inv6;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 && 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] = (1ll * phi[i] * i + phi[i - 1]) % mod;}inv2 = quick_pow(2, mod - 2, mod), inv6 = quick_pow(6, mod - 2, mod); }unordered_map<ll, ll> ans_s;ll S(ll n) {if(n < N) return phi[n];if(ans_s.count(n)) return ans_s[n];ll ans = n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - (l + r) * (r - l + 1) / 2 % mod * S(n / l) % mod + mod) % mod;}return ans_s[n] = ans; }ll solve(ll n) {ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * (r - l + 1) * S(n / l) % mod) % mod;}return (ans + n) % mod; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();ll l = read(), r = read();printf("%lld\n", ((solve(r) - solve(l - 1)) % mod * inv2 % mod + mod) % mod);return 0; }

總結

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