51nod1040 最大公约数之和

求$sum_{i=1}^{n}(i,n)$。n<=1e9。

$sum_{i=1}^{n}(i,n)=sum_{d|n}dsum_{i=1}^{n}[(i,n)=d]=sum_{d|n}dsum_{k=1}^{frac{n}{d}}[(k,frac{n}{d})=1]=sum_{d|n}dvarphi(frac{n}{d})=sum_{d|n}frac{nvarphi(d)}{d}$

枚举因子。搞定。

//#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
//#include<map>
#include<math.h>
//#include<time.h>
//#include<complex>
#include<algorithm>
using namespace std;

int n;
int list[22],num[22],len=0;

#define LL long long
LL ans;
void dfs(int cur,int now,int s)
{
    if (cur>len)
    {
        int p=now;
        for (int i=1;i<=len;i++) if ((s>>(i-1))&1) p=p/list[i]*(list[i]-1);
        ans+=n/now*p; return;
    }
    dfs(cur+1,now,s);
    for (int i=1,tmp=list[cur];i<=num[cur];i++,tmp*=list[cur]) dfs(cur+1,now*tmp,s|(1<<(cur-1)));
}

int main()
{
    scanf("%d",&n);
    int tnt=n;
    for (int i=2;1ll*i*i<=tnt;i++) if (tnt%i==0)
    {
        list[++len]=i;
        while (tnt%i==0) tnt/=i,num[len]++;
    }
    if (tnt) {list[++len]=tnt; num[len]=1;}
    ans=0; dfs(1,1,0); printf("%lld
",ans);
    return 0;
}