P3768 簡單的數(shù)學題

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{m}ijgcd(i,j) \\ =\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^{m}ij(gcd(i,j)=d) \\ =\sum _{d=1}^n{d}^3\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}ij\sum _{k\mid gcd(i,j)}=\mu (k) \\ =\sum _{d=1}^n{d}^3\sum _{k=1}^{\frac{n}{d}}{k}^2\mu (k)\sum _{i=1}^{\frac{n}{kd}}i\sum _{j=1}^{\frac{n}{kd}}j \\ =\sum _{d=1}^n{d}^3\sum _{k=1}^{\frac{n}{d}}{k}^2\mu (k){\left(\frac{?\frac{n}{kd}?(1+?\frac{n}{kd}?)}{2}\right)}^2 \\ 我們假設t=kd \\ =\sum _{t=1}^n{t}^2{\left(\frac{?\frac{n}{t}?(1+?\frac{n}{t}?)}{2}\right)}^2\sum _{k\mid t}\frac{t}{k}\mu (k) \\ =\sum _{t=1}^n{t}^2?(t){\left(\frac{?\frac{n}{t}?(1+?\frac{n}{t}?)}{2}\right)}^2 \\ 所以我們子要可以通過短時間內(nèi)篩選出\sum _{t=1}^n{t}^2?(t)，即可通過數(shù)論分塊來達到要求了。 \end{gathered}$$

$$\begin{gathered} S(n)=\sum _{i=1}^n{i}^2?(i)=\sum _{i=1}^{n}f(i) \\ \sum _{i=1}^n(f?g)(i) \\ =\sum _{i=1}^{n}g(i)S(\frac{n}{i}) \\ g(1)S(n)=\sum _{i=1}^n(f?g)(i)?\sum _{d=2}^{n}g(i)S(\frac{n}{d}) \\ (f?g)(i)=\sum _{d\mid i}f(d)g(\frac{i}{d})=\sum _{d\mid i}{d}^2?(d)g(\frac{i}{d}) \\ 容易想到\sum _{d\mid i}?(d)=i,所以如果可以提出d^2那就完美了，我們可以另g(i)=i^2，上式變成 \\ \sum _{d\mid i}{d}^2?(d)\frac{i^2}{d^2}=i^2\sum _{d\mid i}?(d)=i^3 \\ S(n)=\sum _{i=1}^n{i}^3?\sum _{d=1}^n{d}^{2}S(\frac{n}{d}) \\ =\frac{n^2(n+1{)}^2}{4}?\sum _{d=1}^n{d}^{2}S(\frac{n}{d}) \\ 然后數(shù)論分塊，杜教篩即可得到， \end{gathered}$$

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 = 8e6 + 10;ll phi[N], mod, n, inv4, inv6;int prime[N], cnt;bool st[N];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] + 1ll * i * i % mod * phi[i] % mod) % mod;} }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; }ll calc1(ll x) {x %= mod;return 1ll * x * x % mod * (x + 1) % mod * (x + 1) % mod * inv4 % mod; }ll calc2(ll x) {x %= mod;return x * (x + 1) % mod * (2ll * x + 1) % mod * inv6 % mod; }unordered_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 = calc1(x);for(ll l = 2, r; l <= x; l = r + 1) {r = x / (x / l);ans -= (calc2(r) - calc2(l - 1)) * get_phi(x / l) % mod;ans = (ans % mod + mod) % mod;}return ans_phi[x] = ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);mod = read(), n = read();init();inv4 = quick_pow(4, mod - 2, mod), inv6 = quick_pow(6, mod - 2, mod);ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans += (get_phi(r) - get_phi(l - 1)) % mod * calc1(n / l) % mod;ans = (ans % mod + mod) % mod;}cout << ans << endl;return 0; }

LaTeX公式代碼

\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} ijgcd(i, j)\\ = \sum_{d = 1} ^{n} d \sum_{i = 1} ^{n} \sum_{j = 1} ^{m}ij(gcd(i, j) = d)\\ = \sum_{d = 1} ^{n} d ^ 3 \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}ij \sum_{k \mid gcd(i, j)} = \mu(k)\\ = \sum_{d = 1} ^{n} d ^ 3 \sum_{k = 1} ^{\frac{n}{d}} k ^ 2\mu(k) \sum_{i = 1} ^{\frac{n}{kd}}i \sum_{j = 1} ^{\frac{n}{kd}}j\\ = \sum_{d = 1} ^{n} d ^ 3 \sum_{k = 1} ^{\frac{n}{d}} k ^ 2\mu(k) \left(\frac{\lfloor \frac{n}{kd}\rfloor(1 + \lfloor \frac{n}{kd}\rfloor)}{2}\right) ^ 2\\ 我們假設t = kd\\ =\sum_{t = 1} ^{n} t ^ 2 \left(\frac{\lfloor \frac{n}{t}\rfloor(1 + \lfloor \frac{n}{t}\rfloor)}{2}\right) ^ 2\sum_{k \mid t} \frac{t}{k} \mu(k)\\ = \sum_{t = 1} ^{n} t ^ 2 \phi(t) \left(\frac{\lfloor \frac{n}{t}\rfloor(1 + \lfloor \frac{n}{t}\rfloor)}{2}\right) ^ 2\\ 所以我們子要可以通過短時間內(nèi)篩選出\sum_{t = 1} ^{n} t ^ 2 \phi(t)，即可通過數(shù)論分塊來達到要求了。 S(n) = \sum_{i = 1} ^{n} i ^ 2 \phi(i) = \sum_{i = 1} ^{n} f(i) \\ \sum_{i = 1} ^{n} (f * g)(i) \\ = \sum_{i = 1} ^{n}g(i)S(\frac{n}{i})\\ g(1)S(n) = \sum_{i = 1} ^{n} (f * g)(i) - \sum_{d = 2} ^{n} g(i)S(\frac{n}{d})\\ (f * g)(i) = \sum_{d \mid i} f(d)g(\frac{i}{d}) = \sum_{d \mid i} d ^ 2 \phi(d)g(\frac{i}{d})\\ 容易想到\sum_{d \mid i} \phi(d) = i, 所以如果可以提出d ^ 2那就完美了，我們可以另g(i) = i ^ 2，上式變成\\ \sum_{d \mid i} d ^ 2 \phi(d) \frac{i ^ 2}{d ^ 2} = i ^ 2 \sum_{d \mid i} \phi(d) = i ^ 3 \\S(n) = \sum_{i = 1} ^{n} i ^ 3 - \sum_{d = 1} ^{n} d ^ 2S(\frac{n}{d})\\ = \frac{n ^ 2 (n + 1) ^ 2}{4} - \sum_{d = 1} ^{n} d ^ 2S(\frac{n}{d})\\ 然后數(shù)論分塊，杜教篩即可得到，

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