题解 SP34112 UDIVSUM - The Sum of Unitary Divisors
【分析】
\(\boldsymbol {\sigma^*}\) 的积性容易验证，则仅考虑其在质数幂处的值
\(\displaystyle \boldsymbol {\sigma^*}(p^k)=\sum_{i=0}^kp^i[\gcd(p^i, p^{k-i})=1]=\sum_{i=0}^k p^i[\min(i, k-i)=0]=p^k+[k>0]\)
由于该积性函数具有很好的性质：\(\boldsymbol {\sigma^*}(p)=p+1=\boldsymbol {\sigma}(p)\) ，故考虑 Powerful Number 筛：
令积性函数 \(\boldsymbol h\) 满足 \(\boldsymbol {\sigma^*}=\boldsymbol \sigma*\boldsymbol h\)
\(\begin{aligned} \boldsymbol {\sigma^*}(p^k)&=\sum_{i=0}^k\boldsymbol \sigma(p^i)\boldsymbol h(p^{k-i})&(k>0) \\\\ \boldsymbol {\sigma^*}(p^k)&=\sum_{i=1}^{k-1}\boldsymbol \sigma(p^i)\boldsymbol h(p^{k-i})+\boldsymbol \sigma(p^k)+\boldsymbol h(p^k) \\\\ \boldsymbol h(p^k)&=\boldsymbol {\sigma^*}(p^k)-\boldsymbol \sigma(p^k)-\sum_{i=1}^{k-1}\boldsymbol \sigma(p^i)\boldsymbol h(p^{k-i}) \end{aligned}\)
代入迭代得：
\(\begin{aligned} \boldsymbol h(1)&=1 \\\\ \boldsymbol h(p)&=\boldsymbol {\sigma^*}(p)-\boldsymbol \sigma(p)-\sum_{i=1}^{1-1}\boldsymbol \sigma(p^i)\boldsymbol h(p^{1-i})&=0 \\\\ \boldsymbol h(p^2)&=\boldsymbol {\sigma^*}(p^2)-\boldsymbol \sigma(p^2)-\sum_{i=1}^{2-1}\boldsymbol \sigma(p^i)\boldsymbol h(p^{2-i})&=-p \\\\ \boldsymbol h(p^3)&=\boldsymbol {\sigma^*}(p^3)-\boldsymbol \sigma(p^3)-\sum_{i=1}^{3-1}\boldsymbol \sigma(p^i)\boldsymbol h(p^{3-i})&=0 \end{aligned}\)
继续迭代观察可发现 \(\boldsymbol h(p^k)=0, k\geq 3\) ，可用数学归纳法证明
由此 \(\boldsymbol h(p^k)=\begin{cases} \begin{aligned} -p&, k=2 \\\\0&, k\neq 2 \end{aligned} \end{cases}, k>0\)
且由 Powerful Number 筛，有：
\(\begin{aligned} S(n)&=\sum_{i=1}^n\boldsymbol {\sigma^*}(i) \\\\&=\sum_{i=1}^n\sum_{d\mid i}\boldsymbol h(d)\boldsymbol \sigma({i\over d}) \\\\&=\sum_{d=1}^n \boldsymbol h(d)\sum_{i=1}^n[d\mid i]\boldsymbol \sigma({i\over d}) \\\\&=\sum_{d \in PN} \boldsymbol h(d)\sum_{i=1}^{n/d} \boldsymbol \sigma(i) \\\\&=\sum_{d \in PN} \boldsymbol h(d)S_d(n/d) \end{aligned}\)
而 \(\displaystyle S_d(n)=\sum_{i=1}^n\boldsymbol \sigma(i)=\sum_{i=1}^n \sum_{d\mid i}d=\sum_{d=1}^n d\cdot (n/d)\)
对右边整除分块即可 \(O(\sqrt n)\) 时间内求解 \(S_d(n)\)
总复杂度计算时，考虑到 \(\forall d\in PN\to \exist a, b\in Z^+\wedge d=a^2b^3\)
故 \(\displaystyle T(n)=\sum_{a=1}^{\sqrt n}\sum_{b=1}^{\sqrt[3]{n/a}}\sqrt{n\over a^2b^3}=\int_1^\sqrt n\text da\int_1^{\sqrt[3]{n\over a}}\text db\cdot \sqrt{n\over a^2b^3}=O(\sqrt n\log n)\)
【代码】
还不知道能不能过
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef double db;
#define fi first
#define se second
const int Lim=1e7, MAXN=Lim+10;
int fck[MAXN], prime[MAXN/10], cntprime, lst[MAXN];
ull sigma[MAXN];
inline void init() {
sigma[1]=1;
for(int i=2; i<=Lim; ++i) {
if(!fck[i]) {
fck[i]=prime[++cntprime]=i;
lst[i]=1;
sigma[i]=i+1;
}
for(int j=1; j<=cntprime; ++j)
if( prime[j]*i>Lim ) break;
else if(i%prime[j]) {
fck[ prime[j]*i ]=prime[j];
lst[ prime[j]*i ]=i;
sigma[ prime[j]*i ]=sigma[i]*sigma[ prime[j] ];
}
else {
fck[ prime[j]*i ]=fck[i]*prime[j];
lst[ prime[j]*i ]=lst[i];
if( fck[i]==i ) sigma[ prime[j]*i ]=sigma[i]*prime[j]+1;
else sigma[ prime[j]*i ]=sigma[lst[i]]*sigma[fck[i]*prime[j]];
break;
}
}
for(int i=2; i<=Lim; ++i) sigma[i]+=sigma[i-1];
}
inline ull sum1(ull n) { return (n&1)?((n+1)/2*n):(n/2*(n+1)); }
inline ull sumit(ull m) {
ull res=sigma[min(m, 10000000ull)];
for(ull l=10000000, r, d; l<=m; l=r+1) {
r=m/(d=m/l);
res+=d*(sum1(r)-sum1(l-1));
}
return res;
}
ull dfs(ull n, int flr, ull hn) {
ull m, res=hn*sumit(n);
for(int i=flr+1; i<=cntprime; ++i) {
m=n/prime[i]/prime[i];
if(!m) break;
res+=dfs(m, i, hn*(-prime[i]));
}
return res;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
init();
int T; cin>>T;
ull n;
while(T--&&cin>>n) cout<<dfs(n, 0, 1)<<"\n";
cout.flush();
return 0;
}
