题解 SP34112 UDIVSUM

传送门

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【分析】

(oldsymbol {sigma^*}) 的积性容易验证，则仅考虑其在质数幂处的值

(displaystyle oldsymbol {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：(oldsymbol {sigma^*}(p)=p+1=oldsymbol {sigma}(p)) ，故考虑 Powerful Number 筛：

令积性函数 (oldsymbol h) 满足 (oldsymbol {sigma^*}=oldsymbol sigma*oldsymbol h)

(egin{aligned} oldsymbol {sigma^*}(p^k)&=sum_{i=0}^koldsymbol sigma(p^i)oldsymbol h(p^{k-i})&(k>0) \\ oldsymbol {sigma^*}(p^k)&=sum_{i=1}^{k-1}oldsymbol sigma(p^i)oldsymbol h(p^{k-i})+oldsymbol sigma(p^k)+oldsymbol h(p^k) \\ oldsymbol h(p^k)&=oldsymbol {sigma^*}(p^k)-oldsymbol sigma(p^k)-sum_{i=1}^{k-1}oldsymbol sigma(p^i)oldsymbol h(p^{k-i}) end{aligned})

代入迭代得：

(egin{aligned} oldsymbol h(1)&=1 \\ oldsymbol h(p)&=oldsymbol {sigma^*}(p)-oldsymbol sigma(p)-sum_{i=1}^{1-1}oldsymbol sigma(p^i)oldsymbol h(p^{1-i})&=0 \\ oldsymbol h(p^2)&=oldsymbol {sigma^*}(p^2)-oldsymbol sigma(p^2)-sum_{i=1}^{2-1}oldsymbol sigma(p^i)oldsymbol h(p^{2-i})&=-p \\ oldsymbol h(p^3)&=oldsymbol {sigma^*}(p^3)-oldsymbol sigma(p^3)-sum_{i=1}^{3-1}oldsymbol sigma(p^i)oldsymbol h(p^{3-i})&=0 end{aligned})

继续迭代观察可发现 (oldsymbol h(p^k)=0, kgeq 3) ，可用数学归纳法证明

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由此 (oldsymbol h(p^k)=egin{cases} egin{aligned} -p&, k=2 \\0&, k eq 2 end{aligned} end{cases}, k>0)

且由 Powerful Number 筛，有：

(egin{aligned} S(n)&=sum_{i=1}^noldsymbol {sigma^*}(i) \\&=sum_{i=1}^nsum_{dmid i}oldsymbol h(d)oldsymbol sigma({iover d}) \\&=sum_{d=1}^n oldsymbol h(d)sum_{i=1}^n[dmid i]oldsymbol sigma({iover d}) \\&=sum_{d in PN} oldsymbol h(d)sum_{i=1}^{n/d} oldsymbol sigma(i) \\&=sum_{d in PN} oldsymbol h(d)S_d(n/d) end{aligned})

而 (displaystyle S_d(n)=sum_{i=1}^noldsymbol sigma(i)=sum_{i=1}^n sum_{dmid i}d=sum_{d=1}^n dcdot (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 din PN o exist a, bin Z^+wedge d=a^2b^3)

故 (displaystyle T(n)=sum_{a=1}^{sqrt n}sum_{b=1}^{sqrt[3]{n/a}}sqrt{nover a^2b^3}=int_1^sqrt n ext daint_1^{sqrt[3]{nover a}} ext dbcdot sqrt{nover a^2b^3}=O(sqrt nlog n))

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【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)<<"
";
    cout.flush();
    return 0;
}