题目
我们要求的是
[prodlimits_{i=a}^bprodlimits_{j=1}^i(frac ij)^{lfloorfrac ij
floor}
]
先把它拆开
[prodlimits_{i=a}^bprodlimits_{j=1}^ii^{lfloorfrac ij
floor}(frac1{prodlimits_{i=a}^bprodlimits_{j=1}^ij^{lfloorfrac ij
floor}})
]
对于右边，我们把(jin[1,i])换成(jin[1,b])是没有任何问题的。因为(forall jin(i,b],lfloorfrac ij
floor=0)，相当于多乘了几个(1)。
然后再交换右边的连乘符号
[prodlimits_{i=a}^bprodlimits_{j=1}^ii^{lfloorfrac ij
floor}(frac1{prodlimits_{j=1}^bprodlimits_{i=a}^bj^{lfloorfrac ij
floor}})
]
我们把连乘换成指数的求和
[prodlimits_{i=a}^bi^{sumlimits_{j=1}^ilfloorfrac ij
floor}(frac1{prodlimits_{j=1}^bj^{sumlimits_{i=a}^blfloorfrac ij
floor}})
]
然后容斥一下
[prodlimits_{i=1}^bi^{sumlimits_{j=1}^ilfloorfrac ij
floor}prodlimits_{j=1}^{a-1}j^{sumlimits_{i=1}^{a-1}lfloorfrac ij
floor}(frac1{prodlimits_{j=1}^bj^{sumlimits_{i=1}^blfloorfrac ij
floor}prodlimits_{i=1}^{a-1}i^{sumlimits_{j=1}^ilfloorfrac ij
floor}})
]
[let a=a-1,f(n)=prodlimits_{i=1}^ni^{sumlimits_{j=1}^ilfloorfrac ij
floor},g(n)=prodlimits_{j=1}^nj^{sumlimits_{i=1}^nlfloorfrac ij
floor}
]
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)=sumlimits_{d|n}1,d(n)=sumlimits_{i=1}^nsigma(i)
]
考虑到枚举约数和枚举倍数的等价性
[d(n)=sumlimits_{i=1}^nlfloorfrac ni
floor
]
则
[f(n)=prodlimits_{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)}=prodlimits_{i=1}^ni^{lfloorfrac ni
floor},h(n)=frac{t(n)}{t(n-1)}=prodlimits_{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('
');}
}
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;
}