题目描述
给你\(n\)，求
\[\prod_{i=1}^n{\sigma_0(i)}^{i+\mu(i)}
\]
\]
对\({10}^{12}+39\)取模。
\(\sigma_0(i)\)表示约数个数。
题解
把式子拆成两部分：
\[\prod_{i=1}^n{\sigma_0(i)}^{i+\mu(i)}=\prod_{i=1}^n{\sigma_0(i)}^{i}\prod_{i=1}^n{\sigma_0(i)}^{\mu(i)}
\]
\]
先看前面这部分
\[\begin{align}
S(n)&=\sum_{i=1}^ni=\frac{n(n+1)}{2}\\
\prod_{i=1}^n\sigma_0(i)^i&=\prod_{p}\prod_{k=1}{(k+1)}^{p^kS(\frac{n}{p^k})-p^{k+1}S(\frac{n}{p^{k+1}})}\\
&=\prod_{p\leq \sqrt n}\prod_{k=1}{(k+1)}^{p^kS(\frac{n}{p^k})-p^{k+1}S(\frac{n}{p^{k+1}})}\times \prod_{p>\sqrt n}2^{pS(\frac{n}{p})}
\end{align}
\]
S(n)&=\sum_{i=1}^ni=\frac{n(n+1)}{2}\\
\prod_{i=1}^n\sigma_0(i)^i&=\prod_{p}\prod_{k=1}{(k+1)}^{p^kS(\frac{n}{p^k})-p^{k+1}S(\frac{n}{p^{k+1}})}\\
&=\prod_{p\leq \sqrt n}\prod_{k=1}{(k+1)}^{p^kS(\frac{n}{p^k})-p^{k+1}S(\frac{n}{p^{k+1}})}\times \prod_{p>\sqrt n}2^{pS(\frac{n}{p})}
\end{align}
\]
前面那个式子可以直接计算，后面那个式子可以用min_25筛筛出质数和，然后套上一个数论分块解决。
再看后面那部分：
因为\(i\)的每个质数只出现了一次，所以\(\sigma_0(i)=2^{i\text{的质因子个数}}\)
\[\begin{align}
\prod_{i=1}^n\sigma_0(i)^{\mu(i)}&=\prod_{i=1}^n2^{i\text{的质因子个数}\times \mu(i)}
\end{align}
\]
\prod_{i=1}^n\sigma_0(i)^{\mu(i)}&=\prod_{i=1}^n2^{i\text{的质因子个数}\times \mu(i)}
\end{align}
\]
然后设
\[\begin{align}
F1_{n,j}&=\sum_{i=2}^n[i\text{为质数或}i\text{的最小因子都不小于}p_j]\mu(i)\times i\text{的质因子个数}\\
F2_{n,j}&=\sum_{i=2}^n[i\text{为质数或}i\text{的最小因子都不小于}p_j]\mu(i)\\
F1_{n,j}&=-([p_j,n]\text{之间的质数个数})-\sum_{i\geq j,p_i^2\leq n}(F1_{\frac{n}{p_i},i+1}+F2_{\frac{n}{p_i},i+1})\\
F2_{n,j}&=-([p_j,n]\text{之间的质数个数})-\sum_{i\geq j,p_i^2\leq n}F2_{\frac{n}{p_i},i+1}\\
\end{align}
\]
F1_{n,j}&=\sum_{i=2}^n[i\text{为质数或}i\text{的最小因子都不小于}p_j]\mu(i)\times i\text{的质因子个数}\\
F2_{n,j}&=\sum_{i=2}^n[i\text{为质数或}i\text{的最小因子都不小于}p_j]\mu(i)\\
F1_{n,j}&=-([p_j,n]\text{之间的质数个数})-\sum_{i\geq j,p_i^2\leq n}(F1_{\frac{n}{p_i},i+1}+F2_{\frac{n}{p_i},i+1})\\
F2_{n,j}&=-([p_j,n]\text{之间的质数个数})-\sum_{i\geq j,p_i^2\leq n}F2_{\frac{n}{p_i},i+1}\\
\end{align}
\]
直接上min_25筛就好了。
乘法取模可以用黑科技，或者__int128
时间复杂度：\(O(\frac{n^\frac{3}{4}}{\log n})\)
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<cmath>
#include<functional>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
void sort(int &a,int &b)
{
if(a>b)
swap(a,b);
}
void open(const char *s)
{
#ifndef ONLINE_JUDGE
char str[100];
sprintf(str,"%s.in",s);
freopen(str,"r",stdin);
// sprintf(str,"%s.out",s);
// freopen(str,"w",stdout);
#endif
}
int rd()
{
int s=0,c,b=0;
while(((c=getchar())<'0'||c>'9')&&c!='-');
if(c=='-')
{
c=getchar();
b=1;
}
do
{
s=s*10+c-'0';
}
while((c=getchar())>='0'&&c<='9');
return b?-s:s;
}
void put(int x)
{
if(!x)
{
putchar('0');
return;
}
static int c[20];
int t=0;
while(x)
{
c[++t]=x%10;
x/=10;
}
while(t)
putchar(c[t--]+'0');
}
int upmin(int &a,int b)
{
if(b<a)
{
a=b;
return 1;
}
return 0;
}
int upmax(int &a,int b)
{
if(b>a)
{
a=b;
return 1;
}
return 0;
}
const ll p=1000000000039;
const ll p1=1000000000038;
ll mul(ll a,ll b)
{
return (__int128)a*b%p;
// a%=p;
// b%=p;
// return (a*b-(ll)((long double)a/p*b+1e-8)*p)%p;
}
ll mul1(ll a,ll b)
{
return (__int128)a*b%p1;
// a%=p1;
// b%=p1;
// return (a*b-(ll)((long double)a/p1*b+1e-8)*p1)%p1;
}
ll fp(ll a,ll b){ll s=1;for(;b;b>>=1,a=mul(a,a))if(b&1)s=mul(s,a);return s;}
ll fp1(ll a,ll b){ll s=1;for(;b;b>>=1,a=mul1(a,a))if(b&1)s=mul1(s,a);return s;}
const int M=350010;
const int m=350000;
ll n;
ll sq;
int b[M],pri[M],cnt;
ll f1[M],f2[M],g1[M],g2[M];
void init()
{
for(int i=2;i<=m;i++)
{
if(!b[i])
pri[++cnt]=i;
for(int j=1;j<=cnt&&i*pri[j]<=m;j++)
{
b[i*pri[j]]=1;
if(i%pri[j]==0)
break;
}
}
pri[cnt+1]=m+1;
}
ll s[50];
void gao()
{
for(int i=2;i<=m;i++)
{
f1[i]=(ll)(i+2)*(i-1)/2%p1;
g1[i]=i-1;
}
for(int i=1;n/i>m;i++)
{
f2[i]=((n/i)&1?mul1((n/i-1)/2,(n/i+2)):mul1((n/i+2)/2,(n/i-1)));
g2[i]=n/i-1;
}
for(int i=1;i<=cnt;i++)
{
int j;
ll x1=f1[pri[i]-1];
ll x2=g1[pri[i]-1];
for(j=1;n/j/pri[i]>m&&n/j>=(ll)pri[i]*pri[i];j++)
{
f2[j]=(f2[j]-pri[i]*(f2[j*pri[i]]-x1))%p1;
g2[j]-=g2[j*pri[i]]-x2;
}
for(;n/j>m&&n/j>=(ll)pri[i]*pri[i];j++)
{
f2[j]=(f2[j]-pri[i]*(f1[n/j/pri[i]]-x1))%p1;
g2[j]-=g1[n/j/pri[i]]-x2;
}
for(j=m;j>=(ll)pri[i]*pri[i];j--)
{
f1[j]=(f1[j]-pri[i]*(f1[j/pri[i]]-x1))%p1;
g1[j]-=g1[j/pri[i]]-x2;
}
}
}
pll getf(ll x,int y)
{
if(x<=1||x<pri[y])
return pll();
ll s1=-((x<=m?g1[x]:g2[n/x])-g1[pri[y]-1]);
ll s2=s1;
for(int i=y;i<=cnt&&(ll)pri[i]*pri[i]<=x;i++)
{
pll v=getf(x/pri[i],i+1);
s1=(s1-v.first-v.second)%p1;
s2-=v.second;
}
return pll(s1,s2);
}
ll get(ll x)
{
return x<=m?f1[x]:f2[n/x];
}
ll sum(ll x)
{
return x&1?mul1((x+1)/2,x):mul1(x/2,x+1);
}
void solve()
{
scanf("%lld",&n);
sq=0;
while((sq+1)*(sq+1)<=n)
sq++;
memset(s,0,sizeof s);
gao();
pll res=getf(n,1);
s[2]=(s[2]+res.first)%p1;
for(ll i=1,j;i<=n;i=j+1)
{
j=n/(n/i);
if(j>sq)
s[2]=(s[2]+mul1(get(j)-get(max(i-1,sq)),sum(n/i)))%p1;
}
for(int i=1;i<=cnt&&(ll)pri[i]*pri[i]<=n;i++)
{
ll s1=pri[i],s2=(ll)pri[i]*pri[i];
for(int j=1;s1<=n;s1=s2,s2*=pri[i],j++)
s[j+1]=(s[j+1]+mul1(s1,sum(n/s1))-mul1(s2,sum(n/s2)))%p1;
}
ll ans=1;
for(int i=2;i<=50;i++)
{
s[i]=(s[i]+p1)%p1;
ans=mul(ans,fp(i,s[i]));
}
ans=(ans+p)%p;
printf("%lld\n",ans);
}
int main()
{
open("51nod1965");
init();
int t;
scanf("%d",&t);
while(t--)
solve();
return 0;
}