题解
显然可以O（nlogn）计算
代码
//by 减维 #include<set> #include<map> #include<queue> #include<ctime> #include<cmath> #include<bitset> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define il inline #define rg register #define db double #define mpr make_pair #define maxn 2000005 #define inf (1<<30) #define eps 1e-8 #define pi 3.1415926535897932384626L using namespace std; inline int read() { int ret=0;bool fla=0;char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-'){fla=1;ch=getchar();} while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();} return fla?-ret:ret; } int t,n,num,mu[maxn],pri[maxn],phi[maxn]; ll cnt[maxn],sum[maxn]; bool pd[maxn]; void pre() { phi[1]=1;mu[1]=1; for(int i=2;i<=maxn-5;i++) { if(!pd[i]) pri[++num]=i,phi[i]=i-1,mu[i]=-1; for(int j=1;j<=num&&i*pri[j]<=maxn-5;++j) { pd[i*pri[j]]=1; if(i%pri[j]==0) { phi[i*pri[j]]=phi[i]*pri[j]; break ; } phi[i*pri[j]]=phi[i]*phi[pri[j]]; mu[i*pri[j]]=-mu[i]; } } } il ll gcd(ll x,ll y){return y==0?x:gcd(y,x%y);} ll solve(int x) { memset(cnt,0,sizeof cnt); memset(sum,0,sizeof sum); for(int i=1;i<=x;++i) cnt[phi[i]]++; for(int i=1;i<=x;++i) for(int j=1;i*j<=x;++j) sum[i]+=cnt[i*j]; ll ret=0; for(int d=1;d<=x;++d) if(mu[d]) for(int dd=1;dd*d<=x;++dd) ret+=phi[dd]*mu[d]*sum[d*dd]*sum[d*dd]; return ret; } int main() { t=read(); pre(); while(t--) { n=read(); printf("%lld ",solve(n)); } return 0; }