[sum_{i=0}^{n}inom{n}{i}p^ilfloordfrac{i}{k}
floor\
=sum_{i=0}^{n}inom{n}{i}p^idfrac{i-i\%k}{k}\
=dfrac{1}{k}sum_{i=0}^{n}inom{n}{i}p^i(i-i\%k)\
=dfrac{1}{k}sum_{i=0}^{n}inom{n}{i}p^ii-dfrac{1}{k}sum_{j=0}^{k-1}jsum_{i=0}^{n}[imod k=j]inom{n}{i}p^i
]
(dfrac{1}{k}) 暂且略去。
先看前面，看着很小清新。
[sum_{i=0}^{n}inom{n}{i}p^ii\
=nsum_{i=0}^{n}dfrac{i}{n}inom{n}{i}p^i\
=nsum_{i=0}^{n}inom{n-1}{i-1}p^i\
=npsum_{i=0}^{n-1}inom{n-1}{i}p^i\
=np(p+1)^{n-1}
]
再看后面
[sum_{j=0}^{k-1}jsum_{i=0}^{n}[imod k=j]inom{n}{i}p^i\
=dfrac{1}{k}sum_{j=0}^{k-1}jsum_{i=0}^{n}sum_{t=0}^{k-1}omega_{k}^{t(i-j)}inom{n}{i}p^i\
=dfrac{1}{k}sum_{t=0}^{k-1}sum_{j=0}^{k-1}jomega_{k}^{-tj}sum_{i=0}^{n}inom{n}{i}p^iomega_{k}^{ti}\
=dfrac{1}{k}sum_{t=0}^{k-1}sum_{j=0}^{k-1}jomega_{k}^{-tj}(pomega_{k}^{t}+1)^n\
=dfrac{1}{k}sum_{t=0}^{k-1}(pomega_{k}^{t}+1)^nsum_{j=0}^{k-1}jomega_{k}^{-tj}\
]
后面那个 (omega) 是个等差乘等比的形式，习惯性单独拉出来推。
[S=sum_{j=0}^{k-1}j(omega_{k}^{-t})^j\
omega_{k}^{-t}S=sum_{j=1}^{k}(j-1)(omega_{k}^{-t})^j\
(1-omega_{k}^{-t})S=sum_{j=0}^{k-1}(omega_{k}^{-t})^j-(k-1)omega_{k}^{-tk}-1\
=[k|t]k-(k-1)-1=-k\
S=dfrac{k}{omega_{k}^{-t}-1}
]
哼哼，啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊，一个细节调死人了。这个并不完全是等比数列，开头有 (0) ，我大E了，没有闪，被偷袭了。这个就是第三步 (-1) 的原因。
还要特判 (t=0) 。
#define mod 998244353
inline int qpow(int n,int k){int res=1;for(;k;k>>=1,n=1ll*n*n%mod)if(k&1)res=1ll*n*res%mod;return res;}
int n,p,k,w[1<<20],ans;
signed main(){
n=read(),p=read(),k=read();
w[0]=1,w[1]=qpow(3,(mod-1)/k);
for(int i=2;i<k;++i)w[i]=1ll*w[i-1]*w[1]%mod;
ans=1ll*n*p%mod*qpow(p+1,n-1)%mod;
int res=0;
for(int i=0;i<k;++i){
int x=qpow(1ll*p*w[i]%mod+1,n),y=0;
if(i)y=1ll*k*qpow(w[k-i]-1,mod-2)%mod;
else y=1ll*k*(k-1)/2%mod;
res=(res+1ll*x*y%mod)%mod;
}
res=1ll*res*qpow(k,mod-2)%mod,ans=(ans+mod-res)%mod;
ans=1ll*ans*qpow(k,mod-2)%mod;
cout<<ans<<'
';
}