Luogu P4902 乘积
题目
我们要求的是
\[\prod\limits_{i=a}^b\prod\limits_{j=1}^i(\frac ij)^{\lfloor\frac ij\rfloor}
\]
先把它拆开
\[\prod\limits_{i=a}^b\prod\limits_{j=1}^ii^{\lfloor\frac ij\rfloor}(\frac1{\prod\limits_{i=a}^b\prod\limits_{j=1}^ij^{\lfloor\frac ij\rfloor}})
\]
对于右边，我们把\(j\in[1,i]\)换成\(j\in[1,b]\)是没有任何问题的。因为\(\forall j\in(i,b],\lfloor\frac ij\rfloor=0\)，相当于多乘了几个\(1\)。
然后再交换右边的连乘符号
\[\prod\limits_{i=a}^b\prod\limits_{j=1}^ii^{\lfloor\frac ij\rfloor}(\frac1{\prod\limits_{j=1}^b\prod\limits_{i=a}^bj^{\lfloor\frac ij\rfloor}})
\]
我们把连乘换成指数的求和
\[\prod\limits_{i=a}^bi^{\sum\limits_{j=1}^i\lfloor\frac ij\rfloor}(\frac1{\prod\limits_{j=1}^bj^{\sum\limits_{i=a}^b\lfloor\frac ij\rfloor}})
\]
然后容斥一下
\[\prod\limits_{i=1}^bi^{\sum\limits_{j=1}^i\lfloor\frac ij\rfloor}\prod\limits_{j=1}^{a-1}j^{\sum\limits_{i=1}^{a-1}\lfloor\frac ij\rfloor}(\frac1{\prod\limits_{j=1}^bj^{\sum\limits_{i=1}^b\lfloor\frac ij\rfloor}\prod\limits_{i=1}^{a-1}i^{\sum\limits_{j=1}^i\lfloor\frac ij\rfloor}})
\]
\[let\ a=a-1,f(n)=\prod\limits_{i=1}^ni^{\sum\limits_{j=1}^i\lfloor\frac ij\rfloor},g(n)=\prod\limits_{j=1}^nj^{\sum\limits_{i=1}^n\lfloor\frac ij\rfloor}
\]
则要求的式子就变成了
\[f(b)g(a)(\frac1{f(a)g(b)})
\]
所以如果我们能够以\(O(nlog\ n)\)的复杂的筛出\(f,g\)的话就能解决问题。
首先计算\(f\)
\[let\ \sigma(n)=\sum\limits_{d|n}1,d(n)=\sum\limits_{i=1}^n\sigma(i)
\]
考虑到枚举约数和枚举倍数的等价性
\[d(n)=\sum\limits_{i=1}^n\lfloor\frac ni\rfloor
\]
则
\[f(n)=\prod\limits_{i=1}^ni^{d(i)}
\]
显然其递推式为
\[f(n)=f(n-1)n^{d(n)}
\]
所以我们可以\(O(nlog\ n)\)筛出\(\sigma\)即除数函数，然后前缀和求出,\(d\)，再按递推式求出\(f\)。
注意\(d\)是作为指数存在，所以取模时需要对\(P-1\)取模。
再计算\(g\)
\[let\ t(n)=\frac{g(n)}{g(n-1)}=\prod\limits_{i=1}^ni^{\lfloor\frac ni\rfloor},h(n)=\frac{t(n)}{t(n-1)}=\prod\limits_{d|n}d
\]
显然我们可以\(O(nlog\ n)\)筛出\(h\)即约数积函数，然后做两遍前缀积就可以得到\(g\)。
#include<bits/stdc++.h>
using namespace std;
namespace IO
{
char ibuf[(1<<21)+1],obuf[(1<<21)+1],st[15],*iS,*iT,*oS=obuf,*oT=obuf+(1<<21);
char Get(){return (iS==iT? (iT=(iS=ibuf)+fread(ibuf,1,(1<<21)+1,stdin),(iS==iT? EOF:*iS++)):*iS++);}
void Flush(){fwrite(obuf,1,oS-obuf,stdout),oS=obuf;}
void Put(char x){*oS++=x;if(oS==oT)Flush();}
int read(){int x=0;char ch=Get();while(ch>57||ch<48)ch=Get();while(ch>=48&&ch<=57)x=x*10+(ch^48),ch=Get();return x;}
void write(int x){int top=0;if(!x)Put('0');while(x)st[++top]=(x%10)+48,x/=10;while(top)Put(st[top--]);Put('\n');}
}
using namespace IO;
const int N=1000007,A=1000000,P=993244853;
int inc(int a,int b,int p=P){a+=b;return a>=p? a-p:a;}
int mul(int a,int b){return 1ll*a*b%P;}
int power(int a,int k){int r=1;for(;k;k>>=1,a=mul(a,a))if(k&1)r=mul(a,r);return r;}
int d[N],s[N],f[N];
int main()
{
int i,j,n,a,b;
s[0]=f[0]=1;
for(i=1;i<=A;++i) s[i]=1;
for(i=1;i<=A;++i) for(j=i;j<=A;j+=i) ++d[j],s[j]=mul(s[j],i);
for(i=1;i<=A;++i) d[i]=inc(d[i],d[i-1],P-1),f[i]=mul(f[i-1],power(i,d[i])),s[i]=mul(s[i],s[i-1]);
for(i=1;i<=A;++i) s[i]=mul(s[i],s[i-1]);
for(n=read();n;--n) a=read()-1,b=read(),write(mul(mul(mul(power(f[a],P-2),f[b]),power(s[b],P-2)),s[a]));
return Flush(),0;
}
