毕克大爷的pdf里面的水题
拉格朗日插值法裸题
显然这个函数是个\(k+1\)次多项式,那么插出来把\(n\)往里面带即可
前\(k+1\)项的值可以\(O(n)\)用线性筛处理出来
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// <622F.cpp> - 02/07/17 11:04:48 // This file is created by XuYike's black technology automatically. // Copyright (C) 2015 ChangJun High School, Inc. // I don't know what this program is. #include <iostream> #include <vector> #include <algorithm> #include <cstring> #include <cstdio> #include <cmath> using namespace std; typedef long long lol; int gi(){ int res=0,fh=1;char ch=getchar(); while((ch>'9'||ch<'0')&&ch!='-')ch=getchar(); if(ch=='-')fh=-1,ch=getchar(); while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar(); return fh*res; } const int MAXN=1000010; const int INF=1e9; const int MOD=1e9+7; int qpow(int x,int y){ int res=1; while(y){ if(y&1)res=1ll*res*x%MOD; x=1ll*x*x%MOD; y>>=1; } return res; } int jc[MAXN],nj[MAXN],fj[MAXN],nf[MAXN]; bool np[MAXN]; int pt,pr[MAXN],f[MAXN]; int Pai(){return 1;} template<typename... _> int Pai(int x,_... y){return 1ll*x*Pai(y...)%MOD;} int main(){ int n=gi(),k=gi()+1; jc[0]=1;fj[0]=n; for(int i=1;i<=k;i++){ jc[i]=1ll*jc[i-1]*i%MOD; fj[i]=1ll*fj[i-1]*(n-i)%MOD; } nj[k]=qpow(jc[k],MOD-2); nf[k]=qpow(fj[k],MOD-2); for(int i=k;i;i--){ nj[i-1]=1ll*nj[i]*i%MOD; nf[i-1]=1ll*nf[i]*(n-i)%MOD; } f[1]=1; for(int i=2;i<=k;i++){ if(!np[i]){pr[++pt]=i;f[i]=qpow(i,k-1);} for(int j=1;j<=pt&&i*pr[j]<=k;j++){ np[i*pr[j]]=1; f[i*pr[j]]=1ll*f[i]*f[pr[j]]%MOD; if(!(i%pr[j]))break; } } for(int i=2;i<=k;i++)f[i]=(f[i]+f[i-1])%MOD; if(n<=k){printf("%d",f[n]);return 0;} int ans=0; for(int i=1;i<=k;i++){ int w=Pai(f[i],nj[k-i],nj[i],fj[k],nf[i],i?fj[i-1]:1); if((k-i)&1)ans=(ans-w+MOD)%MOD; else ans=(ans+w)%MOD; } printf("%d",ans); return 0; } |