P3768 简单的数学题（杜教筛）

P3768 简单的数学题

推式子

∑i=1n∑j=1mijgcd(i,j)=∑d=1nd∑i=1n∑j=1mij(gcd(i,j)=d)=∑d=1nd3∑i=1nd∑j=1ndij∑k∣gcd(i,j)=μ(k)=∑d=1nd3∑k=1ndk2μ(k)∑i=1nkdi∑j=1nkdj=∑d=1nd3∑k=1ndk2μ(k)(⌊nkd⌋(1+⌊nkd⌋)2)2我们假设t=kd=∑t=1nt2(⌊nt⌋(1+⌊nt⌋)2)2∑k∣ttkμ(k)=∑t=1nt2ϕ(t)(⌊nt⌋(1+⌊nt⌋)2)2所以我们子要可以通过短时间内筛选出∑t=1nt2ϕ(t)，即可通过数论分块来达到要求了。\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\\ 我们假设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\\ 所以我们子要可以通过短时间内筛选出\sum_{t = 1} ^{n} t ^ 2 \phi(t)，即可通过数论分块来达到要求了。 i=1∑nj=1∑mijgcd(i,j)=d=1∑ndi=1∑nj=1∑mij(gcd(i,j)=d)=d=1∑nd3i=1∑dnj=1∑dnijk∣gcd(i,j)∑=μ(k)=d=1∑nd3k=1∑dnk2μ(k)i=1∑kdnij=1∑kdnj=d=1∑nd3k=1∑dnk2μ(k)(2⌊kdn⌋(1+⌊kdn⌋))2我们假设t=kd=t=1∑nt2(2⌊tn⌋(1+⌊tn⌋))2k∣t∑ktμ(k)=t=1∑nt2ϕ(t)(2⌊tn⌋(1+⌊tn⌋))2所以我们子要可以通过短时间内筛选出t=1∑nt2ϕ(t)，即可通过数论分块来达到要求了。

S(n)=∑i=1ni2ϕ(i)=∑i=1nf(i)∑i=1n(f∗g)(i)=∑i=1ng(i)S(ni)g(1)S(n)=∑i=1n(f∗g)(i)−∑d=2ng(i)S(nd)(f∗g)(i)=∑d∣if(d)g(id)=∑d∣id2ϕ(d)g(id)容易想到∑d∣iϕ(d)=i,所以如果可以提出d2那就完美了，我们可以另g(i)=i2，上式变成∑d∣id2ϕ(d)i2d2=i2∑d∣iϕ(d)=i3S(n)=∑i=1ni3−∑d=1nd2S(nd)=n2(n+1)24−∑d=1nd2S(nd)然后数论分块，杜教筛即可得到，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，我们可以另g(i) = i ^ 2，上式变成\\ \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})\\ 然后数论分块，杜教筛即可得到， S(n)=i=1∑ni2ϕ(i)=i=1∑nf(i)i=1∑n(f∗g)(i)=i=1∑ng(i)S(in)g(1)S(n)=i=1∑n(f∗g)(i)−d=2∑ng(i)S(dn)(f∗g)(i)=d∣i∑f(d)g(di)=d∣i∑d2ϕ(d)g(di)容易想到d∣i∑ϕ(d)=i,所以如果可以提出d2那就完美了，我们可以另g(i)=i2，上式变成d∣i∑d2ϕ(d)d2i2=i2d∣i∑ϕ(d)=i3S(n)=i=1∑ni3−d=1∑nd2S(dn)=4n2(n+1)2−d=1∑nd2S(dn)然后数论分块，杜教筛即可得到，

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;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公式代码

\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\\
我们假设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\\
所以我们子要可以通过短时间内筛选出\sum_{t = 1} ^{n} t ^ 2 \phi(t)，即可通过数论分块来达到要求了。

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，我们可以另g(i) = i ^ 2，上式变成\\
\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})\\
然后数论分块，杜教筛即可得到，