LOJ #6539 奇妙数论题

不想咕太久..就随便找个题更一下

LOJ#6539

---

题意

求题面里那个式子

---

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...

---

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=0;char zf=1;char ch=getchar();
    while(ch!='-'&&!isdigit(ch))ch=getchar();
    if(ch=='-')zf=-1,ch=getchar();
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();return x*zf;
}
void write(ll y){if(y<0)putchar('-'),y=-y;if(y>9)write(y/10);putchar(y%10+48);}
void writeln(const ll y){write(y);putchar('\n');}
int k,m,n,x,y,z,cnt,ans;
#define N 100000
bool pri[N+10];int ss[N+10],phi[N+10];
void init(){
    phi[1]=1;
    for(rt i=2;i<=N;i++){
        if(!pri[i]) ss[++cnt]=i,phi[i]=i-1;
        for(rt j=1;j<=cnt&&i*ss[j]<=N;j++){
            phi[i*ss[j]]=phi[i]*phi[ss[j]];
            pri[i*ss[j]]=1;
            if(i%ss[j]==0){
                phi[i*ss[j]]=phi[i]*ss[j];
                break;
            }
        }
    }
}
int a[100010],sum[100010];
vector<int>ys[100010];
ll calc2(int x){
    ll ret=0;
    for(rt i=x;i<=n;i+=x)sum[a[i]]++;
    for(rt i=1;i<=n;i++){
        int now=0;
        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]]=0;
    return ret;
}
ll calc(int x){
    ll ret=0;
    for(rt i=x;i<=n;i+=x){
        for(auto j:ys[a[i]]){
            ret+=(sum[j]*2+1)*phi[j];
            sum[j]++;
        }
    }
    for(rt i=x;i<=n;i+=x){
        for(auto j:ys[a[i]])sum[j]=0;
    }    
    return ret;
}
int main(){
    init();n=read();
    for(rt i=1;i<=n;i++)a[i]=read();
    for(rt i=1;i<=n;i++)
    for(rt j=i;j<=n;j+=i)ys[j].push_back(i);
    ll ans=0;
    for(rt x=1;x<=n;x++){
        if(x<=10)ans+=1ll*phi[x]*calc2(x);else 
        ans+=1ll*phi[x]*calc(x);
    }
    cout<<ans%1000000007;
    return 0;
}

 

标签：10,ch,LOJ,sum,式子,论题,include,ll,6539
来源： https://www.cnblogs.com/DreamlessDreams/p/10650992.html