【CF622F】The Sum of the k-th Powers

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

// <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;
}

// <622F.cpp> - 02/07/17 11:04:48

// This file is created by XuYike's black technology automatically.

// 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;

}

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