Description
很久很久以前,有一只神犇叫yzy;
很久很久之后,有一只蒟蒻叫lty;
Input
请你读入一个整数N;1<=N<=1E9,A、B模1E9+7;
Output
请你输出一个整数A=sum_{i=1}^N{mu (i^2)};
请你输出一个整数B=sum_{i=1}^N{varphi (i^2)};
Sample Input
1
Sample Output
1
1
Solution
完全不知道第一问是用来干嘛的。。。。反正都是1
第二问,显然,(varphi(i^2)=i imesvarphi(i))
于是就是求,(sum_{i=1}^ni imesvarphi(i))
套用杜教筛的式子,(h=f*g) ,在 (f*g) 中,(sum_{d|n}d imesvarphi(d) imes g(frac{n}{d}))
一种想法是试着把 (frac{n}{d}) 给去掉,那么尝试着将 (g) 定为 (id)
那么 (f*g) 就变成了 (nsum_{d|n}varphi(d)=n^2)
带回杜教筛最后的那个式子,得到,(S(n)=sum_{i=1}i^2-sum_{i=2}^niS(lfloorfrac{n}{i}
floor))
杜教筛求就好了
#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=10000000+10,Mod=1e9+7;
int n,vis[MAXN],prime[MAXN],cnt,phi[MAXN];
ll s[MAXN],inv6;
std::map<int,ll> M;
template<typename T> inline void read(T &x)
{
T data=0,w=1;
char ch=0;
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
x=data*w;
}
template<typename T> inline void write(T x,char ch=' ')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(ch!=' ')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void init()
{
memset(vis,1,sizeof(vis));
vis[0]=vis[1]=0;
phi[1]=1;
for(register int i=2;i<MAXN;++i)
{
if(vis[i])
{
prime[++cnt]=i;
phi[i]=i-1;
}
for(register int j=1;j<=cnt&&i*prime[j]<MAXN;++j)
{
vis[i*prime[j]]=0;
if(i%prime[j])phi[i*prime[j]]=phi[i]*phi[prime[j]];
else
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
}
}
for(register int i=1;i<MAXN;++i)s[i]=(s[i-1]+1ll*i*phi[i]%Mod)%Mod;
}
inline ll qexp(ll a,ll b)
{
ll res=1;
while(b)
{
if(b&1)res=res*a%Mod;
a=a*a%Mod;
b>>=1;
}
return res;
}
inline ll S(int x)
{
if(x<MAXN)return s[x];
if(M.find(x)!=M.end())return M[x];
ll res=0;
for(register int i=2;;)
{
if(i>x)break;
int j=x/(x/i);
(res+=(1ll*(i+j)*(j-i+1)/2)%Mod*S(x/i)%Mod)%=Mod;
i=j+1;
}
return (1ll*x*(x+1)%Mod*(2*x+1)%Mod*inv6%Mod-res+Mod)%Mod;
}
int main()
{
read(n);init();inv6=qexp(6,Mod-2);
printf("1
%lld
",S(n));
return 0;
}