【题解】小猪佩奇学数学
\(\text{Solution:}\)
考虑单位根反演。
\[\frac{1}{k}\sum_{i=0}\binom{n}{i}p^i(i-i\bmod k)
\\
\text{Part1:}
\\
\sum_{i=0}^n \binom{n}{i}p^i i
\\
=np\sum_{i=0}^n \binom{n-1}{i-1}p^{i-1}
\\
=np\sum_{j=0}^{n-1} \binom{n-1}{j}p^j
\\
=np(1+p)^{n-1}
\\
\text{Part2:}
\\
\sum_{i=0}^n \binom{n}{i}p^i(i\bmod k)
\\
=\sum_{i=0}^n \binom{n}{i}p^i\sum_{j=0}^{k-1} j[j\equiv i(\bmod k)]
\\
=\sum_{i=0}^n \binom{n}{i}p^i\sum_{j=0}^{k-1} j[k|i-j]
\\
=\sum_{i=0}^n \binom{n}{i}p^i\sum_{j=0}^{k-1} j\frac{1}{k}\sum_{l=0}^{k-1}\omega_k^{l(i-j)}
\\
=\frac{1}{k}\sum_{j=0}^{k-1} j\sum_{l=0}^{k-1} \omega_k^{-lj}\sum_{i=0}^n \binom{n}{i}p^i\omega_k^{il}
\\
=\frac{1}{k}\sum_{j=0}^{k-1} j\sum_{l=0}^{k-1} \omega_k^{-lj}(1+p\omega_k^l)^n
\\
=\frac{1}{k}\sum_{l=0}^{k-1}(1+p\omega_k^l)^n\sum_{j=0}^{k-1}j\omega_k^{-lj}
\\
=\frac{1}{k}\sum_{l=0}^{k-1}(1+p\omega_k^l)^n\sum_{j=0}^{k-1}j\frac{1}{(\omega_k^l)^j}
\]
后面是等差乘等比数列求和。
复杂度 \(O(k\log n).\)
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod=998244353;
const int g=3;
inline int Add(int x,int y){return (x+y)%mod;}
inline int Mul(int x,int y){return 1ll*x*y%mod;}
inline int qpow(int x,int y){
int res=1;
while(y){
if(y&1)res=Mul(res,x);
x=Mul(x,x);y>>=1;
}
return res;
}
int n,p,k;
int P1,P2;
int num,len=1;
int pwk[1<<21];
inline int getinv(int x){return qpow(x,mod-2);}
signed main(){
scanf("%lld%lld%lld",&n,&p,&k);
P1=Mul(n,Mul(p,qpow((1+p)%mod,n-1)));
while(len<k)len<<=1,num++;
int Wk=qpow(g,(mod-1)/k);
pwk[0]=1;
for(int i=1;i<k;++i)pwk[i]=Mul(pwk[i-1],Wk);
for(int i=0;i<k;++i){
int s1=qpow(Add(1,Mul(p,pwk[i])),n);
int q=pwk[i];
q=getinv(q);
if(q==1)P2=Add(P2,Mul(s1,Mul(k,Mul(k-1,getinv(2)))));
else{
int pt1=Mul(k,qpow(q,k));
pt1=Mul(pt1,getinv(q-1));
int pt2=Mul(q,Add(mod-1,qpow(q,k)));
pt2=Mul(pt2,getinv(Mul(q-1,q-1)));
pt2=mod-pt2;
P2=Add(P2,Mul(Add(pt1,pt2),s1));
}
}
P2=Mul(P2,getinv(k));
P2=mod-P2;
P1=Add(P1,P2);
P1=Mul(P1,getinv(k));
printf("%lld\n",P1);
return 0;
}
