SPOJ

链接

题意：三维平面，找从（0，0，0）看（n，n，n）能看到的点

题解：很明显就是求gcd（i，j，k）==1的（i，j，k）对数，改一下公式即可，记得要算平行坐标轴的三个平面，还有含0的三个坐标

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pii pair<int,int>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-12;
const int N=1000000+10,maxn=400000+10,inf=0x3f3f3f3f;

int mu[N],prime[N],sum[N];
bool mark[N];
void init()
{
    mu[1]=1;
    int cnt=0;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,mu[i]=-1;
        for(int j=1;j<=cnt;j++)
        {
            int t=i*prime[j];
            if(t>N)break;
            mark[t]=1;
            if(i%prime[j]==0){mu[t]=0;break;}
            else mu[t]=-mu[i];
        }
    }
    for(int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i];
}
int main()
{
    init();
    int t,cnt=0;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        ll ans=0;
        for(int i=1,last;i<=n;i=last+1)
        {
            last=n/(n/i);
            ans+=(ll)(sum[last]-sum[i-1])*(n/i)*(n/i)*(n/i);
            ans+=(ll)(sum[last]-sum[i-1])*(n/i)*(n/i)*3;
        }
        printf("%lld
",ans+3);
    }
    return 0;
}
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