20211309 《資訊保安專業導論》第4周學習總結
阿新 • • 發佈:2021-10-16
定義
如果\(f:N\rightarrow R\),滿足對任意互質的正整數\(p,q\),都有\(f(qp)=f(q)f(p)\),則稱f(x)為積性函式
例子:
\(1(n)=1\)
\(id(n)=n\)
\(\epsilon(n)=[n=1],\epsilon(1)=1,\epsilon(n>1)=0\)
\(\phi(n)=1···n中與n互質的個數\)
\(d(n)=n的正因子個數\)
具體實現:
設f為積性函式,假設\(n=p_1^{\alpha 1}p_2^{\alpha 2}···p_k^{\alpha k}\)
則\(f(n)=f(p_1^{\alpha 1})f(p_2^{\alpha 2})···f(p_k^{\alpha k})\)
用質因數分解求f(n)
點選檢視程式碼
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;
}
點選檢視程式碼
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); } } }
應用:
點選檢視程式碼
#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.華華給月月出題
點選檢視程式碼
#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;
}