SPOJ – VLATTICE

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-;
const int N=+,maxn=+,inf=0x3f3f3f3f;int mu[N],prime[N],sum[N];
bool mark[N];
void init()
{
    mu[]=;
    int cnt=;
    for(int i=;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,mu[i]=-;
        for(int j=;j<=cnt;j++)
        {
            int t=i*prime[j];
            if(t>N)break;
            mark[t]=;
            if(i%prime[j]==){mu[t]=;break;}
            else mu[t]=-mu[i];
        }
    }
    for(int i=;i<N;i++)sum[i]=sum[i-]+mu[i];
}
int main()
{
    init();
    int t,cnt=;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        ll ans=;
        for(int i=,last;i<=n;i=last+)
        {
            last=n/(n/i);
            ans+=(ll)(sum[last]-sum[i-])*(n/i)*(n/i)*(n/i);
            ans+=(ll)(sum[last]-sum[i-])*(n/i)*(n/i)*;
        }
        printf("%lld\n",ans+);
    }
    return ;
}
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