BZOJ 2818: Gcd

 1 #include<algorithm>
 2 #include<cstring>
 3 #include<cstdio>
 4 //by NeighThorn
 5 using namespace std;
 6 //大鹏一日同风起，扶摇直上九万里
 7 
 8 const int maxn=10000000+5;
 9 
10 int n,cnt,vis[maxn],miu[maxn],prime[maxn];
11 
12 long long ans=0;
13 
14 signed main(void){
15     scanf("%d",&n);miu[1]=1;cnt=0;
16     for(int i=2;i<=10000000;i++){
17         if(!vis[i])
18             vis[i]=1,prime[++cnt]=i,miu[i]=-1;
19         for(int j=1;j<=cnt&&prime[j]*i<=10000000;j++){
20             vis[i*prime[j]]=1;
21             if(i%prime[j]==0){
22                 miu[i*prime[j]]=0;break;
23             }
24             miu[i*prime[j]]=-miu[i];
25         }
26     }
27     for(int i=2;i<=10000000;i++)
28         miu[i]+=miu[i-1];
29     for(int i=1;i<=cnt&&prime[i]<=n;i++){
30         int lala=n/prime[i];
31         for(int d=1,r;d<=lala;d=r+1){
32             r=lala/(lala/d);
33             if(r>lala)
34                 r=lala;
35             ans+=(long long)(miu[r]-miu[d-1])*(lala/d)*(lala/d);
36         }
37     }
38     printf("%lld\n",ans);
39     return 0;
40 }