HDU-5528 Count a * b(积性函数反演)

HDU-5528 Count a * b

$Description$
$\color {red}{f(n)=\sum_{i=1}^{n}\sum_{j=1}^{n}[n\nmid i\cdot j ]}$
$\color {cyan}{calculate \; g(n)=\sum_{d|n}f(d)}$

$Solution$
$\color {blue}{f(n)=\sum_{i=1}^{n}\sum_{j=1}^{n}[n\nmid i\cdot j ]}$
$\color{orange}{f(n)=n^2-\sum_{i=1}^{n}\sum_{j=1}^{n}[n\mid i\cdot j ]}$
$\color{purple}{令h(n)=\sum_{i=1}^{n}\sum_{j=1}^{n}[n\mid i\cdot j ]}$
$\color {pink}{g(n)=\sum_{d|n}f(d)=\sum_{d|n}d^2-\sum_{d|n}h(d)}$

$h(n)$ 解法一:
$h(n)$ 解法二:

$\color{Magenta}{h(p^k)=(k+1)\cdot p^k-k\cdot p^{k-1}}$
所以可以通过积性函数的性质

$\color{DarkOrange}{F(n)=\sum_{d|n}f(d)=\prod_{p^k\parallel n}(1+f(p)+f(p^2)+..+f(p^k))}$
可以化简得到:

$\color{cyan}{\sum_{d|n}d^2=\prod_{p^k\parallel n}(1+\frac{ p^2\cdot (p^{2\cdot k}-1)}{p^2-1})}$
$\color{blue}{\sum_{d|n}h(d)=\prod_{p^k\parallel n}(k+1)\cdot p^k}$
$\color{red}{g(n)=\prod_{p^k\parallel n}(1+\frac{ p^2\cdot (p^{2\cdot k}-1)}{p^2-1})-\prod_{p^k\parallel n}(k+1)\cdot p^k}$

$Code$

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define sc scanf
#define itn int
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=5e5+5;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
ll mul(ll a,ll b,ll c){ll ans=0;for(;b;b>>=1,a=(a+a)%c)if(b&1)ans=(ans+a)%c;return ans;}
ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=mul(a,a,c))if(b&1)ans=mul(ans,a,c);return ans;}
void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=(a/b)*x;}
int prime[N],tot=0;
bool vis[N]={0};
void pre(){
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
}
ll qk(ll a,ll b){
    ll ans=1;
    for(;b;b>>=1,a=a*a)if(b&1)ans=ans*a;
    return ans;
}
int main(){
    pre();
    int t;cin>>t;
    while(t--){
        int n;sc("%d",&n);
        ull ans=1,res=1;
        for(int i=1;i<=tot&&1ll*prime[i]*prime[i]<=n;i++){
            if(n%prime[i]==0){
                int k=0;
                while(n%prime[i]==0)n/=prime[i],k++;
                ans*=1+1ull*prime[i]*prime[i]*((qk(prime[i],2*k)-1)/(prime[i]*prime[i]-1));
                res*=1ull*(k+1)*qk(prime[i],k);
            }
        }
        if(n>1)ans*=1+1ull*n*n,res*=2ull*n;
        printf("%llu\n",ans-res);
    }   
}

HDU-5528 Count a * b(积性函数 反演)

HDU-5528 Count a * b(积性函数反演)