[largeegin{aligned}
&sum_{i=0}^ninom{n}{i}p^ileft lfloor frac{i}{k}
ight
floor \
=&sum_{i=0}^ninom{n}{i}p^i frac{i-i mod k}{k} \
=&frac{1}{k}left ( sum_{i=0}^ninom{n}{i}p^i i-sum_{i=0}^ninom{n}{i}p^i (i mod k)
ight ) \
end{aligned}
]
先化简前一项：
[largeegin{aligned}
&sum_{i=0}^ninom{n}{i}p^i i \
=&nsum_{i=0}^ninom{n-1}{i-1}p^i \
=&nsum_{i=0}^{n-1}inom{n-1}{i}p^{i+1} \
=&np(p+1)^{n-1}
end{aligned}
]
再化简后一项：
[largeegin{aligned}
&sum_{i=0}^ninom{n}{i}p^i (i mod k) \
=&sum_{d=0}^{k-1}dsum_{i=0}^ninom{n}{i}p^i [k mid i-d] \
=&sum_{d=0}^{k-1}dsum_{i=0}^ninom{n}{i}p^i frac{1}{k}sum_{j=0}^{k-1} omega_k^{ij-dj}\
=&frac{1}{k}sum_{j=0}^{k-1}sum_{i=0}^ninom{n}{i}(p omega_k^j)^isum_{d=0}^{k-1}domega_k^{-dj}\
=&frac{1}{k}sum_{j=0}^{k-1}(p omega_k^j+1)^nsum_{d=0}^{k-1}domega_k^{-dj}\
end{aligned}
]
发现后一项式子形如 (sumlimits_{i=0}^{n-1}ix^i)，应用扰动法得：
[largeegin{aligned}
S=&sum_{i=0}^{n-1}ix^i \
=&sum_{i=0}^{n-1}(i+1)x^{i+1}-nx^n \
=&xsum_{i=0}^{n-1}ix^i+sum_{i=0}^{n-1}x^{i+1}-nx^n \
S=&xS+xfrac{1-x^n}{1-x}-nx^n \
=&frac{xfrac{1-x^n}{1-x}-nx^n}{1-x}
end{aligned}
]
代入得：
[largeegin{aligned}
&frac{1}{k}sum_{j=0}^{k-1}(p omega_k^j+1)^nsum_{d=0}^{k-1}domega_k^{-dj}\
=&frac{1}{k}left(sum_{j=1}^{k-1}(p omega_k^j+1)^nfrac{k}{omega_k^{-j}-1}+(p+1)^nfrac{k(k-1)}{2}
ight)\
end{aligned}
]
然后就能快速计算了。
#include<bits/stdc++.h>
#define maxn 4000010
#define p 998244353
#define g 3
using namespace std;
typedef long long ll;
template<typename T> inline void read(T &x)
{
x=0;char c=getchar();bool flag=false;
while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
if(flag)x=-x;
}
ll n,x,k,w,wn,val;
ll qp(ll x,ll y)
{
ll v=1;
while(y)
{
if(y&1) v=v*x%p;
x=x*x%p,y>>=1;
}
return v;
}
int main()
{
read(n),read(x),read(k),w=1,wn=qp(g,(p-1)/k);
for(int i=0;i<k;++i)
{
ll v=qp(x*w%p+1,n);
if(i) val=(val+v*k%p*qp((qp(w,p-2)-1+p)%p,p-2)%p)%p;
else val=(val+v*(((k-1)*k/2)%p)%p)%p;
w=w*wn%p;
}
printf("%lld",(n*x%p*qp(x+1,n-1)%p-val*qp(k,p-2)%p+p)%p*qp(k,p-2)%p);
return 0;
}