LOJ #6539 奇妙数论题

不想咕太久..就随便找个题更一下

LOJ#6539

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题意

求题面里那个式子

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题解

有一个常用的小式子

$$\sum_{x|a,x|b}\varphi(x)=\gcd(a,b)$$

用这个式子直接对题面的式子进行化简

$$
\begin{aligned}
&\sum_{i=1}^n\sum_{j=1}^n(a_i,a_j)·(i,j)\\
&=\sum_{i=1}^n\sum_{j=1}^n(\sum_{x|i,x|j}\varphi(x))(a_i,a_j)\\
&=\sum_{x=1}^n\varphi(x)\sum_{x|i}\sum_{x|j}(a_i,a_j)
\end{aligned}
$$

枚举x，相当于求一个大小为$ \frac{n}{x}$的集合内两两$ \gcd$的和

再用一次最上面的式子优化

$$
\begin{aligned}
&\sum_{i=1}^n\sum_{j=1}^n\gcd(a_i,a_j)\\
&=\sum_{d=1}^n\varphi(d)(\sum_{i=1}^n[d|a_i])^2
\end{aligned}
$$

预处理每个数的约数，每次暴力计算

复杂度是对的..跑的飞快...

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代码

小范围暴力抢了rk1

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#define rt register int
#define ll long long
using namespace std;
inline ll read(){
    ll x=;char zf=;char ch=getchar();
    while(ch!='-'&&!isdigit(ch))ch=getchar();
    if(ch=='-')zf=-,ch=getchar();
    while(isdigit(ch))x=x*+ch-'',ch=getchar();return x*zf;
}
void write(ll y){if(y<)putchar('-'),y=-y;if(y>)write(y/);putchar(y%+);}
void writeln(const ll y){write(y);putchar('\n');}
int k,m,n,x,y,z,cnt,ans;
#define N 100000
bool pri[N+];int ss[N+],phi[N+];
void init(){
    phi[]=;
    for(rt i=;i<=N;i++){
        if(!pri[i]) ss[++cnt]=i,phi[i]=i-;
        for(rt j=;j<=cnt&&i*ss[j]<=N;j++){
            phi[i*ss[j]]=phi[i]*phi[ss[j]];
            pri[i*ss[j]]=;
            if(i%ss[j]==){
                phi[i*ss[j]]=phi[i]*ss[j];
                break;
            }
        }
    }
}
int a[],sum[];
vector<int>ys[];
ll calc2(int x){
    ll ret=;
    for(rt i=x;i<=n;i+=x)sum[a[i]]++;
    for(rt i=;i<=n;i++){
        int now=;
        for(rt j=i;j<=n;j+=i)now+=sum[j];
        ret+=1ll*phi[i]*now*now;
    }
    for(rt i=x;i<=n;i+=x)sum[a[i]]=;
    return ret;
}
ll calc(int x){
    ll ret=;
    for(rt i=x;i<=n;i+=x){
        for(auto j:ys[a[i]]){
            ret+=(sum[j]*+)*phi[j];
            sum[j]++;
        }
    }
    for(rt i=x;i<=n;i+=x){
        for(auto j:ys[a[i]])sum[j]=;
    }
    return ret;
}
int main(){
    init();n=read();
    for(rt i=;i<=n;i++)a[i]=read();
    for(rt i=;i<=n;i++)
    for(rt j=i;j<=n;j+=i)ys[j].push_back(i);
    ll ans=;
    for(rt x=;x<=n;x++){
        if(x<=)ans+=1ll*phi[x]*calc2(x);else
        ans+=1ll*phi[x]*calc(x);
    }
    cout<<ans%;
    return ;
}