莫比乌斯，欧拉函数题目练习（完结）

$Starttime：2020/11/16Start\ time：2020/11/16$

$Lastupdatetime:2020/11/28Last\ update\ time: 2020/11/28$

$AC22/22AC\ 22 / 22$

解方程

$∑d∣nf(d)σp(nd)=σq(n)f∗σp=σq有σk=∑d∣ndk=idk∗If∗idp∗I=idq∗I∑d∣nμ(d)=μ∗I对上面式子同时卷上一个μf∗idp=idq因为idk是一个完全积性函数，所以idp−1=μ×idpidk∗(μ×idp)=∑d∣ndk×μ(nd)×(nd)k=∑d∣nμ(d)=ϵf=idq∗(μ×idp)f(n)=∑d∣nμ(d)×dp×(nd)qf(1)=1f(n)=nq−np(n∈primes)f(ab)=f(a)×f(b)(gcd(a,b)=1)f(nx)=nq−npn(x−1)q=nxq−np+(x−1)q(n∈primes)然后就可以线性筛了\sum_{d \mid n} f(d) \sigma_{p}(\frac{n}{d}) = \sigma_{q}(n)\\ f * \sigma_{p} = \sigma_{q}\\ 有\sigma_{k} = \sum_{d \mid n} d ^{k} = id_{k} * I\\ f * id_{p} * I = id_{q} * I\\ \sum_{d \mid n} \mu(d) = \mu * I\\ 对上面式子同时卷上一个\mu\\ f * id_{p} = id_{q}\\ 因为id_{k}是一个完全积性函数，所以id_{p} ^{-1} = \mu \times id_{p}\\ id_{k} * (\mu \times id_{p}) = \sum_{d \mid n} d ^k \times \mu(\frac{n}{d}) \times (\frac{n}{d}) ^k = \sum_{d \mid n} \mu(d) = \epsilon\\ f = id_{q} * (\mu \times id_{p})\\ f(n) = \sum_{d \mid n} \mu(d) \times d ^ p \times (\frac{n}{d}) ^{q}\\ f(1) = 1\\ f(n) = n ^ q - n ^ p(n \in primes)\\ f(ab) = f(a) \times f(b) (gcd(a, b) = 1)\\ f(n ^ x) = n ^ q - n ^ p n ^{(x - 1)q} = n ^ {xq} - n ^{p + (x - 1) q} (n \in primes)\\ 然后就可以线性筛了\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e7 + 10, mod = 998244353;ll prime[N], minnp[N], f[N], cnt, p, q, 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() {f[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;f[i] = (quick_pow(i, q) - quick_pow(i, p) + mod) % mod;minnp[i] = 1;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {minnp[i * prime[j]] = minnp[i] + 1;f[i * prime[j]] = f[i / quick_pow(prime[j], minnp[i])] * (quick_pow(prime[j], q * (minnp[i * prime[j]])) - quick_pow(prime[j], p + minnp[i] * q) + mod) % mod;break;}minnp[i * prime[j]] = 1;f[i * prime[j]] = f[i] * f[prime[j]] % mod;}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> n >> p >> q;init();ll ans = 0;for(int i = 1; i <= n; i++) {ans ^= f[i];}cout << ans << "\n";return 0;
}

求和

$\sum_{i = 1} ^{n} \sum_{j = 1} ^{n} gcd(i, j, n)\\ f(n) = \sum_{d \mid n} d \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}[gcd(i, j, \frac{n}{d}) = 1]\\ \sum_{d \mid n} d \sum_{k \mid \frac{n}{d}} \mu(k) \frac{n}{kd} \frac{n}{kd}\\ T = kd\\ \sum_{T \mid n} \frac{n}{T} \frac{n}{T} \sum_{d \mid T} d \mu(\frac{T}{d})\\ \sum_{T \mid n} \frac{n}{T} \frac{n}{T} \phi(T)\\ \sum_{i = 1} ^{n} \sum_{d \mid i} \frac{i}{d} \frac{i}{d} \phi(d)\\ \sum_{d = 1} ^{n} \phi(d) \sum_{i = 1} ^{\frac{n}{d}} i ^ 2\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10;int prime[N], phi[N], cnt, n, mod, inv6;bool st[N];int quick_pow(int a, int n) {int ans = 1;while(n) {if(n & 1) ans = 1ll * ans * a % mod;a = 1ll * 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 = 1; 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] + phi[i - 1]) % mod;}
}unordered_map<int, int> ans_s;int Djs_phi(int n) {if(n < N) return phi[n];if(ans_s.count(n)) return ans_s[n];int ans = 1ll * n * (n + 1) / 2 % mod;for(int l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - 1ll * (r - l + 1) * Djs_phi(n / l) % mod + mod) % mod;}return ans_s[n] = ans;
}int calc(int n) {return 1ll * n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d", &n, &mod);init();inv6 = quick_pow(6, mod - 2);int ans = 0;for(int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * (Djs_phi(r) - Djs_phi(l - 1)) * calc(n / l) % mod + mod) % mod;}printf("%d\n", ans);return 0;
}

算术

$∑i=1n∑j=1mμ(lcm(i,j))∑i=1n∑j=1mμ(igcd(i,j))μ(j)∑i=1n∑j=1mμ(i)μ(j)μ(gcd(i,j))∑d=1nμ(d)∑i=1ndμ(id)∑j=1mdμ(jd)[gcd(i,j)=1]∑d=1nμ(d)∑k=1ndμ(k)∑i=1nkdμ(ikd)∑j=1mkdμ(jkd)T=kd∑T=1n∑i=1nTμ(iT)∑j=1mTμ(jT)∑d∣Tμ(d)×μ(Td)\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \mu(lcm(i, j))\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \mu(\frac{i}{gcd(i, j)}) \mu(j)\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \mu(i) \mu(j) \mu(gcd(i, j))\\ \sum_{d = 1} ^{n} \mu(d) \sum_{i = 1} ^{\frac{n}{d}} \mu(id) \sum_{j = 1} ^{\frac{m}{d}} \mu(jd)[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} \mu(d) \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \sum_{i = 1} ^{\frac{n}{kd}} \mu(ikd) \sum_{j = 1} ^{\frac{m}{kd}} \mu(jkd)\\ T = kd\\ \sum_{T = 1} ^{n} \sum_{i = 1} ^{\frac{n}{T}} \mu(iT) \sum_{j = 1} ^{\frac{m}{T}} \mu(jT) \sum_{d \mid T} \mu(d) \times \mu(\frac{T}{d})\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10;int prime[N], mu[N], g[N], f1[N], f2[N], n, m, cnt;bool st[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 = 1; j <= cnt && 1ll * 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) {g[j] += mu[i] * mu[j / i];}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while(T--) {scanf("%d %d", &n, &m);for(int i = 1; i <= n; i++) {for(int j = i; j <= n; j += i) {f1[i] += mu[j];}}for(int i = 1; i <= m; i++) {for(int j = i; j <= m; j += i) {f2[i] += mu[j];}}int ans = 0;for(int i = 1; i <= n; i++) {ans += f1[i] * f2[i] * g[i];}printf("%d\n", ans);for(int i = 1; i <= n; i++) {f1[i] = 0;}for(int i = 1; i <= m; i++) {f2[i] = 0;}}return 0;
}

Mophues

$∑i=1n∑j=1mf(gcd(i,j))∑d=1nf(d)∑i=1nd∑j=1md[gcd(i,j)=1]∑d=1nf(d)∑k=1ndμ(k)nkdmkd∑T=1nnTmT∑d∣T[f(d)<=p]μ(Td)对于所有输入的p，我们按照从小到大排序，对所有的f(d)从小到大排序用一个树状数组来维护∑d∣T[f(d)<=p]μ(Td)\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} f(gcd(i, j))\\ \sum_{d = 1} ^{n} f(d) \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}}[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} f(d) \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \frac{n}{kd} \frac{m}{kd}\\ \sum_{T = 1} ^{n} \frac{n}{T} \frac{m}{T} \sum_{d \mid T} [f(d) <= p] \mu(\frac{T}{d})\\ 对于所有输入的p，我们按照从小到大排序，对所有的f(d)从小到大排序\\ 用一个树状数组来维护\sum_{d \mid T} [f(d) <= p] \mu(\frac{T}{d})\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;#define int long longtypedef long long ll;const int N = 5e5 + 10;int prime[N], mu[N], cnt;bool st[N];struct F {int f, id;bool operator < (const F &t) const {return f < t.f;}
}a[N];void init() {a[1].id = mu[1] = 1, a[1].f = 0;for(int i = 2; i < N; i++) {a[i].id = i;if(!st[i]) {prime[++cnt] = i;mu[i] = -1;a[i].f = 1;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;a[i * prime[j]].f = a[i].f + 1;if(i % prime[j] == 0) {break;}mu[i * prime[j]] = -mu[i];}}
}struct Ask {int n, m, p, id;bool operator < (const Ask &t) const {return p < t.p;}
}ask[N];int ans[N], tree[N];int lowbit(int x) {return x & (-x);
}void update(int x, int value) {while(x < N) {tree[x] += value;x += lowbit(x);}
}int query(int x) {int ans = 0;while(x) {ans += tree[x];x -= lowbit(x);}return ans;
}signed main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%lld", &T);for(int i = 1; i <= T; i++) {scanf("%lld %lld %lld", &ask[i].n, &ask[i].m, &ask[i].p);ask[i].id = i;if(ask[i].n > ask[i].m) {swap(ask[i].n, ask[i].m);}}sort(a + 1, a + N);sort(ask + 1, ask + 1 + T);int now = 1;for(int i = 1; i <= T; i++) {while(now < N && a[now].f <= ask[i].p) {for(int d = a[now].id; d < N; d += a[now].id) {update(d, mu[d / a[now].id]);}now++;}int res = 0;for(int l = 1, r; l <= ask[i].n; l = r + 1) {r = min(ask[i].n / (ask[i].n / l), ask[i].m / (ask[i].m / l));res += (ask[i].n / l) * (ask[i].m / l) * (query(r) - query(l - 1));}ans[ask[i].id] = res;}for(int i = 1; i <= T; i++) {printf("%lld\n", ans[i]);}return 0;
}

GuGuFishtion(???)

$∑i=1n∑j=1mϕ(ij)ϕ(i)ϕ(j)ϕ(ij)=ϕ(i)ϕ(j)ϕ(gcd(i,j))gcd⁡(i,j)∑i=1n∑j=1mgcd(i,j)ϕ(gcd(i,j))∑d=1ndϕ(d)∑i=1nd∑j=1md[gcd(i,j)=1]∑d=1ndϕ(d)∑k=1ndμ(k)nkdmkd\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \frac{\phi(ij)}{\phi(i) \phi(j)}\\ \phi(ij) = \frac{\phi(i) \phi(j)}{\phi(gcd(i, j))} \gcd(i, j)\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \frac{gcd(i, j)}{\phi(gcd(i, j))}\\ \sum_{d = 1} ^{n} \frac{d}{\phi(d)} \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}}[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} \frac{d}{\phi(d)} \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \frac{n}{kd} \frac{m}{kd}\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 2e6 + 10;int prime[N], phi[N], mu[N], inv[N], n, m, mod, 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() {memset(st, 0, sizeof st);cnt = 0;mu[1] = phi[1] = inv[1] = 1;for(int i = 2; i < N; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;if(!st[i]) {prime[++cnt] = i;mu[i] = -1;phi[i] = i - 1;}for(int j = 1; 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];mu[i * prime[j]] = 0;break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] = (mu[i - 1] + mu[i] + mod) % mod;sum[i] = (sum[i - 1] + 1ll * i * inv[phi[i]] % mod) % mod;}
}ll calc(int n, int m) {if(n > m) swap(n, m);ll ans = 0;for(int l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans = (ans + 1ll * (mu[r] - mu[l - 1]) * (n / l) % mod * (m / l) % mod + mod) % mod;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;scanf("%d", &T);while(T--) {scanf("%d %d %d", &n, &m, &mod);init();ll ans = 0;if(n > m) swap(n, m);for(int l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans = (ans + 1ll * (sum[r] - sum[l - 1]) * calc(n / l, m / l) % mod + mod) % mod;}printf("%lld\n", ans);}return 0;
}

The Boss on Mars

$∑i=1ni4[gcd(i,n)=1]∑d∣nμ(d)d4∑i=1ndi4\sum_{i = 1} ^{n} i ^ 4[gcd(i, n) = 1]\\ \sum_{d \mid n} \mu(d) d ^ 4 \sum_{i = 1} ^{\frac{n}{d}}i ^ 4\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 10, mod = 1e9 + 7;int prime[N], cnt;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}
}int mu(int n) {int sum = 0;for(int i = 1; i <= cnt && 1ll * prime[i] * prime[i] <= n; i++) {if(n % prime[i] == 0) {n /= prime[i];sum++;if(n % prime[i] == 0) {return 0;}}}if(n != 1) sum++;return sum & 1 ? -1 : 1;
}ll calc1(ll n) {ll ans = ans = (1ll * 6 * n % mod * n % mod * n % mod * n % mod * n % mod + 1ll * 15 * n % mod * n % mod * n % mod * n % mod + 1ll * 10 * n % mod * n % mod * n % mod - n + mod) % mod;ans = ans * 233333335 % mod;return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;init();scanf("%d", &T);while(T--) {int n;scanf("%d", &n);ll ans = 0;for(int i = 1; 1ll * i * i <= n; i++) {if(n % i == 0) {int temp = mu(i);if(temp) {ans = (ans + 1ll * temp * i % mod * i % mod * i % mod * i % mod * calc1(n / i) % mod + mod) % mod;}temp = mu(n / i);if(i * i != n && temp) {ans = (ans + 1ll * temp * (n / i) % mod * (n / i) % mod * (n / i) % mod * (n / i) % mod  % mod * calc1(i) % mod + mod) % mod;}}}printf("%lld\n", ans);}return 0;
}

Spare Tire

乱搞后得到一个结论 $a_i = i ^ 2 + i$ 。
$∑i=1n(i2+i)[gcd(i,m)=1]∑d∣mμ(d)d2∑i=1ndi2+∑d∣mμ(d)d∑i=1ndi接下来分解质因子，然后暴力统计即可，复杂度最多不过O(T×1000)\sum_{i = 1} ^{n} (i ^ 2 + i)[gcd(i, m) = 1]\\ \sum_{d \mid m} \mu(d) d^2\sum_{i = 1} ^{\frac{n}{d}} i ^ 2 + \sum_{d \mid m} \mu(d)d \sum_{i = 1} ^{\frac{n}{d}}i\\ 接下来分解质因子，然后暴力统计即可，复杂度最多不过O(T \times 1000)\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 10, mod = 1e9 + 7, inv6 = (mod + 1) / 6;int prime[N], fac[N], tot, cnt, n, m;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}
}ll ans = 0;ll calc1(ll n) {return n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
}ll calc2(ll n) {return n * (n + 1) / 2 % mod;
}void dfs(int sum, int num, int pos) {if(pos > tot) {ll temp = num & 1 ? -1 : 1;ans = ans + temp * sum * sum % mod * calc1(n / sum) % mod + temp * sum * calc2(n / sum) % mod;ans = (ans % mod + mod) % mod;return ;}dfs(sum, num, pos + 1);dfs(sum * fac[pos], num + 1, pos + 1);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();while(scanf("%d %d", &n, &m) != EOF) {tot = 0;for(int i = 1; 1ll * prime[i] * prime[i] <= m; i++) {if(m % prime[i] == 0) {fac[++tot] = prime[i];while(m % prime[i] == 0) {m /= prime[i];}}}if(m != 1) {fac[++tot] = m;}ans = 0;dfs(1, 0, 1);printf("%lld\n", ans);}return 0;
}

整数对

$∑i=1n∑j=1m[p∣i×j]∑i=1n∑j=1m[p∣gcd⁡(i,p)×gcd⁡(j,p)]∑d∣p∑i=1n[gcd(i,p)=d]∑k∣p∑j=1m[gcd(j,p)=k][p∣k×d]∑d∣p∑i=1nd[gcd(i,pd)=1]∑k∣p∑j=1mk[gcd(j,pk)=1][p∣k×d]∑d∣p∑x∣pdμ(x)nxd∑k∣p∑y∣pkμ(y)myk[p∣k×d]先预处理出所有的因子来然后就可以暴力统计答案了复杂度好像是O(T×100×100×100)但是跑不满，最多106左右\sum_{i = 1} ^{n} \sum_{j = 1} ^{m}[p \mid i \times j]\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{m}[p \mid \gcd(i, p) \times \gcd(j, p)]\\ \sum_{d \mid p} \sum_{i = 1} ^{n}[gcd(i, p) = d] \sum_{k \mid p} \sum_{j = 1} ^{m}[gcd(j, p) = k][p \mid k \times d]\\ \sum_{d \mid p}\sum_{i = 1} ^{\frac{n}{d}}[gcd(i, \frac{p}{d}) = 1]\sum_{k \mid p} \sum_{j = 1} ^{\frac{m}{k}}[gcd(j, \frac{p}{k}) = 1][p \mid k \times d]\\ \sum_{d \mid p} \sum_{x \mid \frac{p}{d}} \mu(x) \frac{n}{xd}\sum_{k \mid p} \sum_{y \mid \frac{p}{k}} \mu(y) \frac{m}{yk}[p \mid k \times d]\\ 先预处理出所有的因子来然后就可以暴力统计答案了\\ 复杂度好像是O(T \times 100 \times 100 \times 100)但是跑不满，最多10 ^ 6左右\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 10;int prime[N], mu[N], cnt, n, m, p;bool st[N];vector<int> fac[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 = 1; j <= cnt && 1ll * 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) {fac[j].push_back(i);}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while(T--) {scanf("%d %d %d", &n, &m, &p);ll ans = 0;for(auto d : fac[p]) {ll ans1 = 0;for(auto x : fac[p / d]) {ans1 += mu[x] * (n / x / d);}for(auto k : fac[p]) {ll ans2 = 0;if(1ll * k * d % p) continue;for(auto y : fac[p / k]) {ans2 += mu[y] * (m / y / k);}ans += ans1 * ans2;}}printf("%lld\n", ans);}return 0;
}

gcd

$∑i=1n∑j=1ngcd(xi−1,xj−1)∑i=1n∑j=1nxgcd(i,j)−1∑d=1n(xd−1)∑i=1nd∑j=1nd[gcd(i,j)=1]∑d=1n(xd−1)(2∑i=1nd∑j=1i[gcd(i,j)=1]−1)∑d=1n(xd−1)(2∑i=1ndϕ(i)−1)\sum_{i = 1} ^{n} \sum_{j = 1} ^{n}gcd(x ^ i - 1, x ^ j - 1)\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} x ^{gcd(i, j)} - 1\\ \sum_{d = 1} ^{n} (x ^ d - 1) \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} (x ^ d - 1) \left(2\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{i}[gcd(i, j) = 1] - 1\right)\\ \sum_{d = 1} ^{n} (x ^ d - 1) \left(2\sum_{i = 1} ^{\frac{n}{d}}\phi(i) - 1\right)\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 1e9 + 7;int prime[N], phi[N], cnt;ll sum[N];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 = 1; 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] + phi[i]) % mod;}
}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;
}ll inv(ll x) {return quick_pow(x, mod - 2);
}ll calc(ll x, int l, int r) {if(x == 1) return 0;return (quick_pow(x, l) * inv(x - 1) % mod * (quick_pow(x, r - l + 1) - 1) % mod - (r - l + 1) + 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);int T;init();scanf("%d", &T);while(T--) {ll x, n, ans = 0;scanf("%lld %lld", &x, &n);for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + calc(x, l, r) * (2ll * phi[n / l] - 1) % mod) % mod;}printf("%lld\n", ans);}return 0;
}

Clarke and math

$\sum_{i_1 \mid i} \sum_{i_2 \mid i_2} \sum_{i_3 \mid i_2} \dots\sum_{i_k \mid i_{k - 1}}f(i_k)\\ 迪利克雷卷积的定义有f * g(n) = \sum_{d \mid n} f(d) f(\frac{n}{d})\\ I * f = \sum_{d \mid n} f(d)\\ 所以g(i) = (I_1 * I_2 * I_3 \dots I_{k - 1} * I_{k}) * f\\ 所以只要做一下迪利克雷卷积的快速幂就好了\\ 整体复杂度n \log n \log k\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 10, mod = 1e9 + 7;int n, k;struct matrix {ll a[N];void init() {for(int i = 1; i <= n; i++) {a[i] = 0;}}void print() {for(int i = 1; i <= n; i++) {printf("%d%c", a[i], i == n ? '\n' : ' ');}}
}a, ans, I;matrix operator * (const matrix & a, const matrix & b) {matrix ans;ans.init();for(int i = 1; i <= n; i++){for(int j = i; j <= n; j += i) {ans.a[j] = (ans.a[j] + a.a[i] * b.a[j / i] % mod) % mod;}}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;scanf("%d", &T);while(T--) {scanf("%d %d", &n, &k);for(int i = 1; i <= n; i++) {scanf("%d", &a.a[i]);I.a[i] = 1;}ans.init();ans.a[1] = 1;while(k) {if(k & 1) ans = ans * I;I = I * I;k >>= 1;}ans = ans * a;ans.print();}return 0;
}

Code

$10000\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n}gcd(i, j) \times(gcd(i, j) - 1) \times cnt_i \times cnt_j\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} gcd(i, j) ^ 2 \times cnt_i \times cnt_j\\ \sum_{d = 1} ^{n} d ^ 2 \sum_{i = 1} ^{\frac{n}{d}} cnt_{id} \sum_{j = 1} ^{\frac{n}{d}} cnt_{jd}[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} d ^ 2 \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \left(\sum_{i = 1} ^{\frac{n}{kd}}cnt_{ikd} \right) ^ 2\\ T = kd\\ \sum_{T = 1} ^{n} \left(\sum_{i = 1} ^{\frac{n}{T}} cnt_{iT} \right) ^ 2 \sum_{d \mid T} d ^ 2 \mu(\frac{T}{d})\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} gcd(i, j) \times cnt_i \times cnt_j同理可得\\ \sum_{T = 1} ^{n} \left(\sum_{i = 1} ^{\frac{n}{T}} cnt_{iT} \right) ^ 2\sum_{d \mid T} d \mu(\frac{T}{d})\\ \sum_{T = 1} ^{n} \left(\sum_{i = 1} ^{\frac{n}{T}}cnt_{iT} \right) ^ 2 \sum_{d \mid T} (d ^ 2 - d) \mu(\frac{T}{d})\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e4 + 10, mod = 1e4 + 7;int prime[N], mu[N], num[N], cnt, n;ll g[N], f[N];bool st[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 = 1; j <= cnt && 1ll * 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) {f[j] = (f[j] + (1ll * i * i - i) * mu[j / i] % mod + 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);init();while(scanf("%d", &n) != EOF) {memset(g, 0, sizeof g);memset(num, 0, sizeof num);for(int i = 1; i <= n; i++) {int x;scanf("%d", &x);num[x]++;}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {g[i] = (g[i] + num[j]) % mod;}}ll ans = 0;for(int i = 1; i < N; i++) {ans =( ans + g[i] * g[i] % mod * f[i]) % mod;}printf("%lld\n", ans);}return 0;
}

AT5200 [AGC038C] LCMs

$∑i=0n−2∑j=i+1n−1lcm(ai,aj)更改下标从1开始∑i=1n−1∑j=i+1nlcm(ai,aj)(∑i=1n∑j=1nlcm(ai,aj)−∑i=1nlcm(ai,ai))×inv2∑i=1n∑j=1nlcm(ai,aj)−∑i=1nain=1e6∑i=1n∑j=1ni×jgcd(i,j)×cnti×cntj∑d=1nd∑i=1ndi×cntid∑j=1ndj×cntjd[gcd(i,j)=1]∑d=1nd∑k=1ndk2μ(k)(∑i=1nkdi×cntikd)2T=kd∑T=1nT(∑i=1nTi×cntiT)2∑k∣Tkμ(k)\sum_{i = 0} ^{n - 2} \sum_{j = i + 1} ^{n - 1}lcm(a_i, a_j)\\ 更改下标从1开始\\ \sum_{i = 1} ^{n - 1} \sum_{j = i + 1} ^{n} lcm(a_i, a_j)\\ (\sum_{i = 1} ^{n} \sum_{j = 1} ^{n} lcm(a_i, a_j) - \sum_{i = 1} ^{n} lcm(a_i, a_i)) \times inv2\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} lcm(a_i, a_j) - \sum_{i = 1} ^{n} a_i\\ n = 1e6\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} \frac{i \times j}{gcd(i, j)} \times cnt_i \times cnt_j\\ \sum_{d = 1} ^{n} d \sum_{i = 1} ^{\frac{n}{d}}i \times cnt_{id} \sum_{j = 1} ^{\frac{n}{d}} j\times cnt_{jd}[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} d \sum_{k = 1} ^{\frac{n}{d}} k ^ 2 \mu(k) \left(\sum_{i = 1} ^{\frac{n}{kd}} i \times cnt_{ikd} \right) ^ 2\\ T = kd\\ \sum_{T = 1} ^{n}T \left(\sum_{i = 1} ^{\frac{n}{T}} i \times cnt_{iT} \right) ^ 2 \sum_{k \mid T} k \mu(k)$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 998244353;ll sum[N], c[N], n, m, all;bool st[N];int prime[N], mu[N], cnt;ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;n >>= 1;a = a * a % mod;}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] + i * mu[i] % mod + 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);init();scanf("%lld", &n), m = 0;for(int i = 1; i <= n; i++) {ll x;scanf("%lld", &x);all = (all + x) % mod;m = max(x, m);c[x]++;}ll ans = 0;for(ll t = 1; t <= m; t++) {ll res = 0;for(ll i = 1; i <= m / t; i++) {res = (res + 1ll * i * c[i * t] % mod) % mod;}ans = (ans + t * res % mod * res % mod * sum[t] % mod) % mod;}printf("%lld\n", ((ans - all) * quick_pow(2, mod - 2, mod) % mod + mod) % mod);return 0;
}

Dirichlet k -th root

$\sum_{d \mid n}f(d) g(\frac{n}{d})\\ g = f ^ k = f * f * f \dots * f\\ give\ g\ solve\ f\\ f ^ {k ^ {\frac{1}{k}}} = f = g ^{\frac{1}{k}} = g ^{inv(k)}\\ 然后迪利克雷卷积快速幂即可\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 10, mod = 998244353;int n, k;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;
}ll inv(int x) {return quick_pow(x, mod - 2);
}struct matrix {ll a[N];void init() {memset(a, 0, sizeof a);}matrix operator * (const matrix &t) const {matrix ans;ans.init();for(int i = 1; i <= n; i++) {for(int j = i; j <= n; j += i) {ans.a[j] = (ans.a[j] + a[i] * t.a[j / i] % mod) % mod;}}return ans;}
}a, ans;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d", &n, &k);k = inv(k);for(int i = 1; i <= n; i++) {scanf("%lld", &a.a[i]);}ans.a[1] = 1;while(k) {if(k & 1) ans = ans * a;a = a * a;k >>= 1;}for(int i = 1; i <= n; i++) {printf("%lld%c", ans.a[i], i == n ? '\n' : ' ');}return 0;
}

GCD of Divisors

$\sum_{d \mid n} gcd(d, \frac{n}{d})\\ \sum_{i = 1} ^{n} \sum_{d \mid i} gcd(d, \frac{i}{d})\\ \sum_{d = 1} ^{n} \sum_{d \mid i} gcd(d, \frac{i}{d})\\ \sum_{d = 1} ^{n} \sum_{i = 1} ^{\frac{n}{d}} gcd(d, i)\\ \sum_{d = 1} ^{n} \sum_{i = 1} ^{\frac{n}{d}} \sum_{k \mid gcd(d, i)} \phi(k)\\ \sum_{d = 1} ^{n} \sum_{k \mid d} \phi(k) \frac{n}{kd}\\ \sum_{k = 1} ^{n} \phi(k) \sum_{d = 1} ^{\frac{n}{k}} \frac{n}{k ^ 2 d}\\ 设f(n) = \sum_{i = 1} ^{n} \frac{n}{i}\\ 原式子 = \sum_{k = 1} ^{k \times k <= n} \phi(k) f(\frac{n}{k ^ 2})\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 4e7 + 10;ll prime[N], phi[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 = 1; 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);}}
}ll calc(ll n) {ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans += (r - l + 1) * (n / l);}return ans;
}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 n, ans = 0;scanf("%lld", &n);for(int i = 1; 1ll * i * i <= n; i++) {ans += phi[i] * calc(n / i / i);}cout << ans << "\n";return 0;
}

Absolute Math

$the\ prime\ n\ have，c(1) = 0, c(2) = 1, c(4) = 1, c(6) = 2 \\ f(n) = \sum_{d \mid n} \mu(d) ^ 2 = 2 ^{c(n)} \\ f(ab) = \frac{f(a) \times f(b)}{f(gcd(a, b))}\\ \sum_{i = 1} ^{m} \frac{f(n) \times f(i)}{f(gcd(n, i))}\\ f(n) \sum_{d \mid n} \frac{1}{f(d)} \sum_{i = 1} ^{\frac{m}{d}} f(id)[gcd(i, \frac{n}{d}) = 1]\\ f(n) \sum_{d \mid n} \frac{1}{f(d)} \sum_{k \mid \frac{n}{d}} \mu(k) \sum_{i = 1} ^{\frac{m}{kd}} f(ikd)\\ T = kd\\ f(n) \sum_{T \mid n} \sum_{i = 1} ^{\frac{m}{T}}f(iT) \sum_{d \mid T} \frac{\mu(\frac{T}{d})}{f(d)}\\ F(m, n) = \sum_{i = 1} ^{m} f(in), g(n) = \sum_{d \mid n} \frac{\mu(\frac{n}{d})}{f(d)}\\ F(m, n) = f(n) \sum_{T \mid n} F(\frac{m}{T}, T) g(T)\\ g = \mu * \frac{1}{f}，是一个积性函数\\ g(1) = 1\\ g(p) = \frac{\mu(p)}{f(1)} + \frac{\mu(1)}{f(p)} = -1 + \frac{1}{2} = -\frac{1}{2}\\ g(p ^ x)[x >= 2] = \frac{\mu(1)}{f(p ^ x)} + \frac{\mu(p)}{f(p ^{x - 1})} = \frac{1}{2} - \frac{1}{2} = 0\\ 然后递归求解，卡卡常\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e7 + 10, mod = 1e9 + 7, inv2 = mod + 1 >> 1;int prime[N], sum[N], g[N], f[N], cnt;bool st[N];void init() {f[1] = g[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;f[i] = 2;g[i] = mod - inv2;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {f[i * prime[j]] = f[i];break;}f[i * prime[j]] = f[i] * 2;g[i * prime[j]] = 1ll * g[i] * g[prime[j]] % mod;}}for(int i = 1; i < N; i++) {sum[i] = (f[i] + sum[i - 1]) % mod;}
}ll solve(int m, int n) {if(m == 0) return 0;if(m == 1) return f[n];if(n == 1) return sum[m];ll ans = 0;//发现m <= 20的时候是较优的……if(m <= 20 && 1ll * n * m < N) {for(int i=1;i<=m;i++)ans = (ans + f[i * n]) % mod;return ans;}for(int i = 1; 1ll * i * i <= n; i++) {if(n % i == 0){if(g[i]){ans = (ans + solve(m / i, i) * g[i] % mod) % mod;}if(i * i !=n && g[n / i]){ans = (ans + solve(m / (n / i), n / i)*g[n / i]) % mod;}}}return ans * f[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();int T;scanf("%d", &T);while(T--) {int n, m;scanf("%d %d", &n, &m);printf("%lld\n", solve(m, n));}return 0;
}
/*杭电的测评机比自己的老年机跑的还慢……
*/

Counting Divisors

$∑i=1nd(ik)caculte:∑i=lrd(ik)d(n)是一个积性函数d(1)=1d(p)=2d(pk)=k+1然后用区间质数筛就可以得到答案了。\sum_{i = 1} ^{n} d(i ^ k)\\ caculte:\sum_{i = l} ^{r} d(i ^ k)\\ d(n)是一个积性函数\\ d(1) = 1\\ d(p) = 2\\ d(p ^ k) = k + 1\\ 然后用区间质数筛就可以得到答案了。$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 998244353;int prime[N], cnt;ll a[N], num[N], l, r, k;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while(T--) {scanf("%lld %lld %lld", &l, &r, &k);for(int i = 0; i <= r - l; i++) a[i] = l + i, num[i] = 1;for(int i = 1; 1ll * prime[i] * prime[i] <= r; i++) {ll st = l / prime[i] * prime[i];if(st < l) st += prime[i];for(ll j = st; j <= r; j += prime[i]) {ll res = 0;while(a[j - l] % prime[i] == 0) {res++;a[j - l] /= prime[i];}num[j - l] = num[j - l] * (res * k % mod + 1) % mod;}}for(int i = 0; i <= r - l; i++) {if(a[i] != 1) {num[i] = num[i] * (k + 1) % mod;}}ll ans = 0;for(int i = 0; i <= r - l; i++) {ans = (ans + num[i]) % mod;}printf("%lld\n", ans);}return 0;
}

Master of Phi

$\sum_{d \mid n} \phi(d) \frac{n}{d}\\ n = \prod_{i = 1} ^{n} p_i ^{a_i}\\ f(n) = (\phi * id)(n)\\ f = \phi * id\\ f * I = \phi * id * I = id * id = id \times \sigma(id)\\ f = (id \times \sigma(id) ) * \mu = \sum_{d \mid n} \mu(d) \times \frac{n}{d} \times \sigma(\frac{n}{d})\\ f(p ^ k) = p ^{k - 1}\times(pk + p - k)\\ 因为f(n)是一个积性函数，所以只要算出每一个质数项然后直接相乘即可\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int mod = 998244353;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;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T, m;scanf("%d", &T);while(T--) {ll ans = 1, p, k;scanf("%d", &m);for(int i = 1; i <= m; i++) {scanf("%lld %lld", &p, &k);ans = (p * k % mod + p - k + mod) % mod * ans % mod * quick_pow(p, k - 1) % mod;}printf("%lld\n", ans);}return 0;
}

Just a Math Problem

$f(k)equalthenumberofprimefactorsinkg(k)=2f(k)=∑d∣kμ(d)2∑i=1ng(i)∑i=1n∑d∣iμ(d)2μ(d)2=∑k2∣dμ(k)∑i=1n∑d∣i∑k2∣dμ(k)∑k=1k×k≤nμ(k)∑k2∣d∑d∣i∑k=1k×k≤nμ(k)∑a=1nk2∑b=1nak2∑k=1k×k≤nμ(k)∑i=1nk2nik2f(k)\ equal\ the\ number\ of\ prime\ factors\ in\ k\\ g(k) = 2 ^{f(k)} = \sum_{d \mid k} \mu(d) ^ 2 \\ \sum_{i = 1} ^{n} g(i)\\ \sum_{i = 1} ^{n} \sum_{d \mid i} \mu(d) ^ 2\\ \mu(d) ^ 2 = \sum_{k ^ 2 \mid d} \mu(k)\\ \sum_{i = 1} ^{n} \sum_{d \mid i} \sum_{k ^ 2 \mid d} \mu(k)\\ \sum_{k = 1} ^{k \times k \leq n} \mu(k) \sum_{k ^ 2 \mid d} \sum_{d \mid i}\\ \sum_{k = 1} ^{k \times k \leq n} \mu(k) \sum_{a = 1} ^{\frac{n}{k ^ 2}} \sum_{b = 1} ^{\frac{n}{a k ^ 2}}\\ \sum_{k = 1} ^{k \times k \leq n} \mu(k) \sum_{i = 1} ^{\frac{n}{k ^ 2}} \frac{n}{i k ^ 2}\\$

#include<cstdio>
typedef long long ll;
const int N=1000010,P=1000000007;
int T,C,tot,p[N/10],i,j,ans,f[N];char mu[N],v[N];ll n;
inline int F(ll n){if(n<N)if(f[n])return f[n];ll t=0;for(ll i=1,j;i<=n;i=j+1)j=n/(n/i),t+=n/i*(j-i+1);t%=P;if(n<N)f[n]=t;return t;
}
int main(){for(mu[1]=1,i=2;i<N;i++){if(!v[i])mu[i]=-1,p[tot++]=i;for(j=0;j<tot&&i*p[j]<N;j++){v[i*p[j]]=1;if(i%p[j])mu[i*p[j]]=-mu[i];else break;}}for(scanf("%d",&T);T--;printf("Case #%d: %d\n",++C,(ans+P)%P)){scanf("%I64d",&n);for(ans=0,i=1;i<=n/i;i++)if(mu[i])ans=(ans+F(n/i/i)*mu[i])%P;}return 0;
}

D. Winter is here

$\sum k[gcd(a_{i1}, a_{i2}, a_{i3} \dots a_{i_{k - 1}}, a_{k}) = n]\\ ans = \sum_{i = 2} ^{n} i f(i)\\ F(n) = \sum k[n \mid a_{i1}, n \mid a_{i2}, n \mid _{ai3} \dots n \mid a_{i_{k - 1}}, n \mid a_{i_k}]\\ cnt_n表示是n的倍数的数字的个数\\ = \sum_{i = 1} ^{cnt_n} i C_{cnt_n} ^{i} = cnt_n \times 2 ^{cnt_n - 1}\\ F(d) = \sum_{d \mid n} f(n)\\ f(d) = \sum_{d \mid n} F(n) \mu(\frac{n}{d})\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 1e9 + 7;int prime[N], mu[N], num[N], f[N], cnt, n;vector<int> fac[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 = 1; j <= cnt && 1ll * 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 = 2 * i; j < N; j += i) {num[i] += num[j];}if(num[i]) {num[i] = 1ll * num[i] * quick_pow(2, num[i] - 1) % mod;}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {f[i] = (f[i] + 1ll * num[j] * mu[j / i] % mod + 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);cin >> n;for(int i = 1; i <= n; i++) {int x;cin >> x;num[x]++;}init();ll ans = 0;for(int i = 2; i < N; i++) {ans = (ans + 1ll * i * f[i] % mod) % mod;}cout << ans << "\n";return 0;
}

Battlestation Operational

$∑i=1n∑j=1i⌈ij⌉[gcd(i,j)=1]∑i=1n∑j=1i(ij+1)[gcd(i,j)=1]−∑i=1n∑j=1i[gcd(i,j)=1,j∣i]∑i=1n∑j=1i(ij+1)[gcd(i,j)=1]−n∑i=1n∑j=1iij[gcd(i,j)=1]+∑i=1n∑j=1i[gcd(i,j)=1]−n∑i=1n∑j=1iij[gcd(i,j)=1]+∑i=1ϕ(i)−n∑i=1n∑j=1iij[gcd(i,j)=1]∑d=1nμ(d)∑i=1nd∑j=1iij求解∑i=1n∑j=1iij=∑i=1n∑j=inji可以发现这个东西具有区间分块性质所以可以利用差分然后求前缀和得到，具体实现看代码整体复杂度O(nlog⁡n+T×n)\sum_{i = 1} ^{n} \sum_{j = 1} ^{i} \lceil \frac{i}{j} \rceil [gcd(i, j) = 1]\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{i}( \frac{i}{j} + 1)[gcd(i, j) = 1] - \sum_{i = 1} ^{n}\sum_{j = 1} ^{i}[gcd(i, j) = 1, j \mid i]\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{i}(\frac{i}{j} + 1)[gcd(i, j) = 1] - n\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{i} \frac{i}{j}[gcd(i, j) = 1] + \sum_{i = 1} ^{n} \sum_{j = 1} ^{i} [gcd(i, j) = 1] - n\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{i} \frac{i}{j}[gcd(i, j) = 1] + \sum_{i = 1} \phi(i) - n\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{i} \frac{i}{j}[gcd(i, j)= 1]\\ \sum_{d = 1} ^{n} \mu(d) \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{i} \frac{i}{j}\\ 求解\sum_{i = 1} ^{n} \sum_{j = 1} ^{i} \frac{i}{j} = \sum_{i = 1} ^{n} \sum_{j = i} ^{n} \frac{j}{i}\\ 可以发现这个东西具有区间分块性质\\所以可以利用差分然后求前缀和得到，具体实现看代码\\ 整体复杂度O(n \log n + T \times \sqrt n)\\$

/*Author : lifehappy
*/	
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 1e9 + 7;int prime[N], phi[N], mu[N], f[N], cnt, n;bool st[N];void init() {mu[1] = phi[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;mu[i] = -1;phi[i] = i - 1;}for(int j = 1; 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);mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {int l = j, r = j + i;f[l] = (f[l] + (j / i)) % mod;if(r < N) f[r] = (f[r] - (j / i) + mod) % mod;}}for(int i = 1; i < N; i++) {phi[i] = (phi[i] + phi[i - 1]) % mod;mu[i] = (mu[i] + mu[i - 1] + mod) % mod;f[i] = (f[i] + f[i - 1]) % mod;}for(int i = 1; i < N; i++) {f[i] = (f[i] + f[i - 1]) % 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();while(scanf("%d", &n) != EOF) {ll ans = 0;for(int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * (mu[r] - mu[l - 1]) * f[n / l] % mod + mod) % mod;}ans = (ans + phi[n] - n + mod) % mod;printf("%lld\n", ans);}return 0;
}

RXD and math

$∑i=1nkμ2(i)×nki\sum_{i = 1} ^{n ^ k} \mu ^ 2(i) \times \sqrt{\frac{n ^ k}{i}}\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int mod = 1e9 + 7;int quick_pow(ll a, ll n) {a %= mod;ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}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, k, cas = 1;while(cin >> n >> k) {printf("Case #%d: %d\n", cas++, quick_pow(n, k));}return 0;
}

GCD?LCM!

$\sum_{i = 1} ^{n} \sum_{j = 1} ^{n} [lcm(i, j) + gcd(i, j) \geq n]\\ F(n) = n ^ 2 - \sum_{i = 1} ^{n} \sum_{j = 1} ^{n}[lcm(i, j) + gcd(i, j) < n)]\\ F(n) - F(n - 1) = n ^ 2 - (n - 1) ^ 2 - \sum_{i = 1} ^{n} \sum_{j = 1} ^{n}[lcm(i, j) + gcd(i, j) < n] + \sum_{i = 1} ^{n - 1} \sum_{j = 1} ^{n - 1}[lcm(i, j) + gcd(i, j) < n - 1]\\ F(n) - F(n - 1) = 2 \times n - 1 - \sum_{i = 1} ^{n - 1} \sum_{j = 1} ^{n - 1} [lcm(i, j) + gcd(i, j) = n - 1]\\ g(n) = \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} [lcm(i, j) + gcd(i, j) = n]\\ g(n) = \sum_{d \mid n} \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}[ijd + d = n ][gcd(i, j) = 1]\\ = \sum_{d \mid n} \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}[ij = \frac{n}{d} - 1][gcd(i, j = 1)]\\ \sum_{d \mid n} \sum_{i = 1} ^{\frac{n}{d} - 1} \sum_{j = 1} ^{\frac{n}{d} - 1}[ij = \frac{n}{d} - 1][gcd(i, j) = 1]\\ k(n) = \sum_{i = 1} ^{n} \sum_{j = 1} ^{n}[ij = n][gcd(i, j) = 1]\\ k(1) = 1, k(p) = 2,k (p ^ k) = 2，且为积性函数所以可以质数筛得到\\ 然后再通过一次埃筛得到g(n)，即可O(n)得到F(n)\\$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 258280327;ll prime[N], k[N], primes[N], g[N], f[N], cnt, n;bool st[N];void init() {k[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {primes[i] = i;prime[++cnt] = i;k[i] = 2;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {k[i * prime[j]] = k[i / primes[i]] * 2;primes[i * prime[j]] = primes[i] * prime[j];break;}k[i * prime[j]] = k[i] * 2;primes[i * prime[j]] = prime[j];}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {g[j] = (g[j] + k[j / i - 1]) % mod;}}for(int i = 1; i < N; i++) {f[i] = (f[i - 1] + 2ll * i - 1 - g[i - 1] + mod) % mod;}for(int i = 1; i < N; i++) {f[i] = (f[i - 1] + f[i]) % 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();int T;scanf("%d", &T);while(T--) {scanf("%d", &n);printf("%lld\n", f[n]);}return 0;
}

莫比乌斯，欧拉函数题目练习（完结）

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