洛谷$P1390$ 公约数的和 欧拉函数

正解:欧拉函数

解题报告:

传送门$QwQ$

首先显然十分套路地变下形是趴

$\begin{align*}&=\sum_{i=1}^n\sum_{j=1}^n gcd(i,j)\\&=\sum_{i=1}^n\sum_{j=1}^n\sum_{d=1}^{min(i,j)} [gcd(i,j)==d]\cdot d\\&=\sum_{d=1}^{n}d\cdot \sum_{i=1}^n\sum_{j=1}^n [gcd(i,j)==d]\\\end{align*}$

然后就欧拉函数做呗?直接戳我简要总结里常见套路第一条,就能$O(n)$做了$QwQ$

(说下昂,这题里其实是无序的,所以最后用$phi$的时候就可以直接$i\cdot \phi(i)$鸭$QwQ$

其实本来还有一种方法的但因为某不便透露的原因被删了$kk$

 

#include
using namespace std;
#define il inline
#define fi first
#define sc second
#define gc getchar()
#define mp make_pair
#define int long long
#define P pair
#define ri register int
#define rc register char
#define rb register bool
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];i;i=edge[i].nxt)

const int N=2000000+10;
int n,phi[N],sum[N],pr[N],pr_cnt,as;
bool is_pr[N];

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
il void pre()
{
    phi[1]=1;
    rp(i,2,N-10)
    {
        if(!is_pr[i])pr[++pr_cnt]=i,phi[i]=i-1;;sum[i]=sum[i-1]+phi[i];
        rp(j,1,pr_cnt)
        {
            if(pr[j]*i>N-10)break;;is_pr[pr[j]*i]=1;
            if(!(i%pr[j])){phi[i*pr[j]]=phi[i]*pr[j];break;}
            phi[i*pr[j]]=phi[i]*phi[pr[j]];
        }
    }
}

signed main()
{
    //freopen("1390.in","r",stdin);freopen("1390.out","w",stdout);
    pre();n=read();rp(i,1,n/2)as+=i*sum[n/i];
    printf("%lld\n",as);
    return 0;
}

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