P4213 【模板】杜教篩（Sum）

套路推式子

$$\begin{gathered} 求s(n)=\sum _{i=1}^{n}f(i) \\ \sum _{i=1}^n(f?g)(i) \\ =\sum _{i=1}^n\sum _{d\mid i}f(d)g(\frac{i}{d}) \\ =\sum _{d=1}^n\sum _{i=1}^{?\frac{n}{d}?}f(i)g(d) \\ =\sum _{d=1}^{n}g(d)S(?\frac{n}{d}?) \\ =g(1)S(n)+\sum _{d=2}^{n}g(d)S(?\frac{n}{d}?) \\ 則有g(1)S(n)=\sum _{i=1}^n(f?g)(i)?\sum _{d=2}^{n}g(d)S(?\frac{n}{d}?) \end{gathered}$$
莫比烏斯函數求和
$$\begin{gathered} 對S(n)=\sum _{i=1}^N\mu (i) \\ \sum _{i=1}^n(I?\mu )(i) \\ =\sum _{d=1}^{n}S(\frac{n}{d}) \\ 1=S(n)+\sum _{d=2}^{n}S(\frac{n}{d}) \\ S(n)=1?\sum _{d=2}^{n}S(\frac{n}{d}) \end{gathered}$$

歐拉函數求和
$$\begin{gathered} 對S(n)=\sum _{i=1}^n?(i) \\ \sum _{i=1}^n(I??)(i) \\ =\sum _{d=1}^{n}S(\frac{n}{d}) \\ =S(n)+\sum _{d=2}^n?(\frac{n}{d}) \\ S(n)=\frac{n(n+1)}{2}?\sum _{d=2}^n?(\frac{n}{d}) \end{gathered}$$

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 = 5e6 + 10;ll phi[N], mu[N]; int prime[N], cnt = 0; bool st[N];void init() {phi[1] = mu[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 = 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);mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) mu[i] += mu[i - 1], phi[i] += phi[i - 1]; }unordered_map<int, ll> ans_phi; unordered_map<int, ll> ans_mu;ll get_phi(ll x) {if(x < N) return phi[x];if(ans_phi[x]) return ans_phi[x];ll ans = x * (x + 1) >> 1;for(ll l = 2, r; l <= x; l = r + 1) {r = x / (x / l);ans -= (r - l + 1) * get_phi(x / l);}return ans_phi[x] = ans; }ll get_mu(ll x) {if(x < N) return mu[x];if(ans_mu[x]) return ans_mu[x];int ans = 1;for(ll l = 2, r; l <= x; l = r + 1) {r = x / (x / l);ans -= (r - l + 1) * get_mu(x / l);}return ans_mu[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);init();int T = read();while(T--) {ll n = read();printf("%lld %lld\n", get_phi(n), get_mu(n));}return 0; }

總結

以上是生活随笔為你收集整理的杜教筛模板（P4213 【模板】杜教筛（Sum））的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。