[Luogu5320][BJOI2019]堪破神机(DP+斯特林数)

时间:2024-09-27 15:05:14

https://www.cnblogs.com/cjyyb/p/10747543.html

特征方程+斯特林反演化简式子,要注意在模998244353意义下5没有二次剩余,所以每个数都要用$a+b\sqrt{5}$的形式表示,运算类似复数。

斯特林反演的几个用法:

1.下降幂转幂:连续求和时可以通过等比数列求和公式加速。

2.幂转下降幂:类似自然数幂和地用有限微积分加速,或帮助设计DP状态。

 #include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std; const int N=,mod=,i2=,i5=,i6=; ll V,L,R;
int T,m,K,S[N][N],C[N][N]; struct num{ int a,b; }z;
num operator +(const num &a,const num &b){ return (num){(a.a+b.a)%mod,(a.b+b.b)%mod}; }
num operator -(const num &a,const num &b){ return (num){(a.a-b.a+mod)%mod,(a.b-b.b+mod)%mod}; }
num operator *(const num &a,const num &b){ return (num){int((1ll*a.a*b.a+V*a.b*b.b)%mod),int((1ll*a.b*b.a+1ll*a.a*b.b)%mod)}; }
num operator *(const num &a,int b){ return (num){int(1ll*a.a*b%mod),int(1ll*a.b*b%mod)}; } int ksm(int a,ll b){
int res=;
for (; b; a=1ll*a*a%mod,b>>=)
if (b & ) res=1ll*res*a%mod;
return res;
} num ksm(num a,ll b){
num res=(num){,};
for (; b; a=a*a,b>>=)
if (b & ) res=res*a;
return res;
} num Inv(num a){ return (num){a.a,(mod-a.b)%mod}*ksm((1ll*a.a*a.a-V*a.b*a.b%mod+mod)%mod,mod-); } int main(){
freopen("calc.in","r",stdin);
freopen("calc.out","w",stdout);
scanf("%d%d",&T,&m);
if (m==) V=; else V=;
while (T--){
scanf("%lld%lld%d",&L,&R,&K); S[][]=;
rep(i,,K) rep(j,,i) S[i][j]=(S[i-][j-]-1ll*S[i-][j]*(i-)%mod+mod)%mod;
rep(i,,K) C[i][]=;
rep(i,,K) rep(j,,i) C[i][j]=(C[i-][j]+C[i-][j-])%mod;
num a,b,A,B; ll len=R-L+;
if (m==) L++,R++,a=(num){i2,i2},b=(num){i2,mod-i2},A=(num){,i5},B=(num){,mod-i5};
else R>>=,L=(L+)>>,a=(num){,},b=(num){,mod-},A=(num){i2,i6},B=(num){i2,mod-i6};
int ans=; ll t=R-L;
rep(i,,K){
num res=z;
rep(j,,i){
num s=ksm(A,j)*ksm(B,i-j)*C[i][j];
num p=ksm(ksm(a,L),j)*ksm(ksm(b,L),(i-j));
num q=ksm(a,j)*ksm(b,i-j);
num w=p*((ksm(q,t+)-(num){,})*Inv(q-(num){,}));
if (q.a== && q.b==) w=p*((R-L+)%mod);
res=res+s*w;
}
ans=(ans+1ll*res.a*S[K][i])%mod;
}
int Ans=1ll*ans*ksm(len%mod,mod-)%mod;
rep(i,,K) Ans=1ll*Ans*ksm(i,mod-)%mod;
printf("%d\n",Ans);
}
return ;
}