51nod1040 最大公约数之和

求$\sum_{i=1}^{n}(i,n)$。n<=1e9。

$\sum_{i=1}^{n}(i,n)=\sum_{d|n}d\sum_{i=1}^{n}[(i,n)=d]=\sum_{d|n}d\sum_{k=1}^{\frac{n}{d}}[(k,\frac{n}{d})=1]=\sum_{d|n}d\varphi(\frac{n}{d})=\sum_{d|n}\frac{n\varphi(d)}{d}$

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[],num[],len=;

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

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