思路：杜教筛
提交：(2)次
错因：(sum)函数处取模出错
题解：
首先第一问是智商检测题：(sum_{i=1}^n mu(i^2))显然为(1)
第二问其实是跟杜教筛板子那篇里面说的似的：
(f=varphi(i^2)=varphi(i)cdot i)
(S(n)=sum_{i=1}^{n}f(i))
配式子：
(id^2=fcdot id)
所以式子就成了
(1cdot S(n)=sum_{i=1}^{n}i^2-sum_{i=2}^{n}icdot S(frac{n}{i}))
#include<cstdio>
#include<iostream>
#include<map>
#define ll long long
#define R register int
using namespace std;
namespace Luitaryi {
template<class I> inline I g(I& x) { x=0; register I f=1;
register char ch; while(!isdigit(ch=getchar())) f=ch=='-'?-1:f;
do x=x*10+(ch^48); while(isdigit(ch=getchar())); return x*=f;
} const int N=5000000,M=1e9+7,Inv=166666668;
int n,cnt,p[N+10];
ll phi[N+10]; bool v[N+10];
map<int,ll> mem;
inline void PRE() { v[1]=phi[1]=1;
for(R i=2;i<=N;++i) {
if(!v[i]) p[++cnt]=i,phi[i]=i-1;
for(R j=1;j<=cnt&&i*p[j]<=N;++j) {
v[i*p[j]]=true;
if(i%p[j]==0) {
phi[i*p[j]]=phi[i]*p[j]; break;
} phi[i*p[j]]=phi[i]*(p[j]-1);
}
} for(R i=1;i<=N;++i) phi[i]=(1ll*phi[i]*i%M+phi[i-1])%M;
}
inline ll sum(ll x) {return 1ll*x*(x+1)%M*(2*x+1)%M*Inv%M;}
inline ll query(ll x) {return 1ll*x*(x+1)/2%M;}
inline ll solve(int n) {
if(n<=N) return phi[n];
if(mem.count(n)) return mem[n];
register ll ret=sum(n);
for(R l=2,r;l<=n;l=r+1)
r=min(n/(n/l),n),ret-=(query(r)-query(l-1))%M*solve(n/l)%M,ret=(ret+M)%M;
return mem[n]=ret;
}
inline void main() {PRE(); g(n); printf("1
%lld
",solve(n));}
} signed main() {Luitaryi::main(); return 0;}
2019.08.2324
7776