积性函数

定义

如果\(f:N\rightarrow R\),满足对任意互质的正整数\(p,q\),都有\(f(qp)=f(q)f(p)\),则称f(x)为积性函数

例子：

\(1(n)=1\)
\(id(n)=n\)
\(\epsilon(n)=[n=1],\epsilon(1)=1,\epsilon(n>1)=0\)
\(\phi(n)=1···n中与n互质的个数\)
\(d(n)=n的正因子个数\)

具体实现：

设f为积性函数，假设\(n=p_1^{\alpha 1}p_2^{\alpha 2}···p_k^{\alpha k}\)
则\(f(n)=f(p_1^{\alpha 1})f(p_2^{\alpha 2})···f(p_k^{\alpha k})\)

用质因数分解求f(n)

点击查看代码

int solve(int x){
    int ans=1;
    for(int i=2;i<=sqrt(x);i++){
        int cnt=0;
        while(x%i==0){x/=i,cnt++;}
        ans*=calc_f(i,cnt);
    }
    if(x>1)ans*=calc_f(x,1);
    return ans;
}

用欧拉筛求f(1)···f(n)

点击查看代码

void init(){
    f[1]=1;
    for(int i=2;i<=maxn;i++){
        if(!is[i])prime[++tot]=i,cnt[i]=1,f[i]=calc_f(i,1);
        for(int j=1;j<=tot && i<=maxn/prime[j];j++){
            is[i*prime[j]]=1;
            if(i%prime[j]==0){
                cnt[i*prime[j]]=cnt[i]+1;
                f[i*prime[j]]=f[i]/calc_f(prime[j],cnt[i])*calc_f(prime[j],cnt[i]+1);
                break;
            }
            cnt[i*prime[j]]=1;
            f[i*prime[j]]=f[i]*calc_f(prime[j],1);
        }
    }
}

应用：

1. 求n的正因子个数

点击查看代码

#include<functional>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<queue>
#include<deque>
#define ll long long 
using namespace std;
const int maxn=10000000+101;
const int MOD=1000000007;
const ll inf=2147483647;
int read(){
    int x=0,f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
    return x*f;
}
int q,f[maxn],cnt[maxn];
int tot,prime[maxn],is[maxn];

int calc_f(int x,int i){return i+1;}
void init(){
    f[1]=1;
    for(int i=2;i<=maxn;i++){
        if(!is[i])prime[++tot]=i,cnt[i]=1,f[i]=calc_f(i,1);
        for(int j=1;j<=tot && i<=maxn/prime[j];j++){
            is[i*prime[j]]=1;
            if(i%prime[j]==0){
                cnt[i*prime[j]]=cnt[i]+1;
                f[i*prime[j]]=f[i]/calc_f(prime[j],cnt[i])*calc_f(prime[j],cnt[i]+1);
                break;
            }
            cnt[i*prime[j]]=1;
            f[i*prime[j]]=f[i]*calc_f(prime[j],1);
        }
    }
}
int main(){
    q=read();init();
    for(int i=1;i<=q;i++){
        int x=read();printf("%d\n",f[x]);
    }
    return 0;
}

2.华华给月月出题

点击查看代码

#include<functional>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<queue>
#include<deque>
#define ll long long 
using namespace std;
const int maxn=14000000+101;
const int MOD=1e9+7;
const int inf=2147483647;
int read(){
    int x=0,f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
    return x*f;
}
ll f[maxn];
int n,tot,prime[maxn],is[maxn];
ll M(ll x){return (x%MOD+MOD)%MOD;}
ll power(ll x,ll y){
    ll ans=1;
    while(y){
        if(y&1)ans=ans*x%MOD;
        y>>=1;x=x*x%MOD;
    }
    return ans%MOD;
}
ll calc_f(int x,int i){return power((ll)x,(ll)i);}
void init(){
    f[1]=1ll;
    for(int i=2;i<=n;i++){
        if(!is[i])prime[++tot]=i,f[i]=calc_f(i,n)%MOD;
        for(int j=1;j<=tot && i<=n/prime[j];j++){
            is[i*prime[j]]=1;f[i*prime[j]]=f[i]*f[prime[j]]%MOD;
            if(i%prime[j]==0)break;
        }
    }
}
int main(){
    n=read();init();ll ans=0;
    for(int i=1;i<=n;i++)ans^=f[i];printf("%lld",ans);
    return 0;
}

3.E. Bash Plays with Functions 题解