标签：== 约数个数 范围 const begin int its redo else
给定一个正整数\(N(N\le2^{31}-1)\)
求
\(\begin{aligned} ans_1=\sum_{i=1}^n\varphi(i) \end{aligned}\)
\(\begin{aligned} ans_2=\sum_{i=1}^n \mu(i) \end{aligned}\)
一共T+1行
第1行为数据组数T(T<=10)
第2~T+1行每行一个非负整数N，代表一组询问
一共T行，每行两个用空格分隔的数ans1,ans2
6
1
2
8
13
30
2333 1 1
2 0
22 -2
58 -3
278 -3
1655470 2\(N \leq 2^{31}\)
对于任意函数f，g，有\(\begin{aligned} h(i) = \sum_{d|i}f(d)*g(\frac{n}{d})\end{aligned}\)
h即为f和g的卷积
1、\(i(n) = 1\)
2、\(id(n) = n\)
3、\(e(n)=\left\{\begin{aligned}1\ \ \ n = 1 \\ 0 \ \ \ n \neq 1\end{aligned}\right.\)
4、欧拉函数\(\varphi(n)\)
5、懵逼钨丝函数\(\mu(n)=\left\{\begin{aligned}1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 1 \\ (-1)^k \ \ \ n由k个不同质数相乘得到\\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 其它情况\end{aligned}\right.\)
6、\(\sigma(n)=n的约数和\)
7、\(d(n)=n的约数个数\)
1、\(i*\mu = e\)
2、\(e*a=a\)
3、\(\mu * id= \varphi\)
4、\(i*id=\sigma\)
5、\(i*i=d\)
6、\(i*\varphi=id\)
已知\(f(i)\)
用来求\(\begin{aligned}\sum_{i = 1}^n f(i)\end{aligned},n\leq 2^{31}\)
定义\(h(i)=(f*g)(i)=\begin{aligned}\sum_{d|i}f(d)*g(\frac{i}{d})\end{aligned}\)
\(\displaystyle\sum_{i=1}^nh(i)\)
用定义展开
\(=\displaystyle\sum_{i=1}^n\sum_{d|i}g(d)f\left(\frac i d\right)\)
d的范围也是【1.n】的，所以改成枚举d，找它的倍数，这个式子是在求和，找全了就行
\(=\displaystyle \sum_{d=1}^ng(d)\sum_{d|i}f\left(\frac i d \right)\)
把后面变一下
\(=\displaystyle \sum_{d=1}^ng(d)\sum_{i=1}^{\left\lfloor\frac n d \right \rfloor}f( i)\)
然后
\(=\displaystyle \sum_{i=1}^ng(i)S\left(\left\lfloor\frac n i\right\rfloor\right)\)
所以
\(\displaystyle \sum_{i=1}^nh(i)=\sum_{i=1}^ng(i)S\left(\left\lfloor\frac n i\right\rfloor\right)\)
有一个好像没用的式子
\(\displaystyle g(1)S(n)=\sum_{i=1}^ng(i)S\left(\left\lfloor\frac n i\right\rfloor\right)-\sum_{i=2}^ng(i)S\left(\left\lfloor\frac n i\right\rfloor\right)\)
上式把后面移项就成恒等式了
我们把右面第一项用刚刚的结论换走
\(\displaystyle g(1)S(n)=\sum_{i=1}^nh(i)-\sum_{i=2}^ng(i)S\left(\left\lfloor\frac n i\right\rfloor\right)\)
这。。是个递归式
就没了
对于S的递归，用数列分块
一般的h和g都很好求（构造）
对于本题来说
\(i*\varphi=id\)
所以对于\(\varphi\)
\(\displaystyle S(n)=\frac{n*(n+1)}{2}-\sum_{i=2}^nS\left(\left\lfloor\frac n i\right\rfloor\right)\)
刚刚有\(i*\mu=e\)
所以
\(\displaystyle S(n)=1-\sum_{i=2}^nS\left(\left\lfloor\frac n i\right\rfloor\right)\)
没了。。。
把前\(4*10^6\)的东西线性筛一下
最后的复杂度\(O(n^{\frac{2}{3}})\)不会证
#include <bits/stdc++.h>
typedef long long LL;
const int maxn = 4e6;
const int maxx = 4e6 + 10;
int in() {
char ch; int x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
return x * f;
}
bool vis[maxx];
LL phi[maxx];
int mu[maxx], pri[maxx], tot;
std::map<int, LL> P;
std::map<int, int> M;
void predoit() {
phi[1] = mu[1] = 1LL;
for(int i = 2; i <= maxn; i++) {
if(!vis[i]) {
pri[++tot] = i;
phi[i] = i - 1;
mu[i] = -1;
}
for(int j = 1; j <= tot && i * pri[j] <= maxn; j++) {
vis[i * pri[j]] = true;
if(i % pri[j] == 0) {
phi[i * pri[j]] = phi[i] * pri[j];
mu[i * pri[j]] = 0;
break;
}
else {
phi[i * pri[j]] = phi[i] * (pri[j] - 1);
mu[i * pri[j]] = -mu[i];
}
}
}
for(int i = 2; i <= maxn; i++) {
phi[i] += phi[i - 1];
mu[i] += mu[i - 1];
}
}
LL workphi(int now)
{
if(now <= maxn) return phi[now];
if(P.count(now)) return P[now];
LL ans = now * (now + 1LL) / 2;
for(int i = 2, lst; i <= now; i = lst + 1) {
lst = now / (now / i);
ans -= 1LL * (lst - i + 1LL) * workphi(now / i);
}
return P[now] = ans;
}
int workmu(int now)
{
if(now <= maxn) return mu[now];
if(M.count(now)) return M[now];
int ans = 1;
for(int i = 2, lst; i <= now; i = lst + 1) {
lst = now / (now / i);
ans -= workmu(now / i) * (lst - i + 1);
}
return M[now] = ans;
}
int main() {
predoit();
for(int T = in(); T --> 0;) {
int n = in();
printf("%lld %d\n", workphi(n), workmu(n));
}
return 0;
}标签：== 约数个数 范围 const begin int its redo else
原文地址：https://www.cnblogs.com/olinr/p/10165706.html
