簡單題 加強版

由簡單版中，我們已經推出了

\[\sum_{d=1}^n\mu^2(d)d^{k+1}\sum_{t=1}^{\lfloor {\frac{n}{d}} \rfloor}\mu(t)t^k\sum_{i=1}^{\lfloor {\frac{n}{dt}} \rfloor}\sum_{j=1}^{\lfloor {\frac{n}{dt}} \rfloor}(i+j)^k \]

我們設\(T=td\),則
設\(S(x)=\sum_{i=1}^x\sum_{j=1}^x(i+j)^k\)
$\sum_{d=1}^n \mu ^2 (d)d\mu (T/d)\sum_{T=1}^n T^k S(n/T)$

點選檢視程式碼

#include <bits/stdc++.h>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ll int
#define ull unsigned long long
#define lid (rt<<1)
#define rid (rt<<1|1)
// #define endl '\n'
//#define int long long
#define pb push_back
// #pragma comment(linker, “/STACK:512000000,512000000”) 
using namespace std;
const int N = 1e7+5;
int f[N*2],mu[N*2],F[N*2],pr[N*2],tot,p[N*2],s[N*2];
bool is[N*2];
int n,k;
ll qpow(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1)ans=ans*a;
		b>>=1;
		a=a*a;
	}
	return ans;
}
void getMu()
{
	f[1]=1;mu[1]=1;p[1]=1;
	for(int i=2;i<=n*2;i++)
	{
		if(!is[i])
		{
			pr[++tot]=i;p[i]=qpow(i,k);
			mu[i]=-1;f[i]=i-1;
		}
		for(int j=1;j<=tot&&i*pr[j]<=2*n;j++)
		{
			p[i*pr[j]]=p[i]*p[pr[j]];
			is[i*pr[j]]=1;
			if(i%pr[j]==0)
			{
				int q=i/pr[j];
				if(q%pr[j])f[i*pr[j]]=-pr[j]*f[q];
				break;
			}
			mu[i*pr[j]]=-mu[i];
			f[i*pr[j]]=f[i]*(pr[j]-1);
		}
	}
	for(int i=2;i<=2*n;i++)f[i]=f[i-1]+p[i]*f[i],p[i]+=p[i-1];
	for(int i=2;i<=2*n;i++)p[i]+=p[i-1];
}
ll calc(ll n)
{
	return p[n<<1]-p[n]*2;
}
int main()
{
	speed();
	// freopen("in.in","r",stdin);
	// freopen("out.out","w",stdout);
	int T;
	cin>>T>>n>>k;
	getMu();
	while(T--)
	{
		ll tn;
		cin>>tn;
		int l=1,r,ans=0;
		while(l<=tn)
		{
			r=tn/(tn/l);
			ans+=(f[r]-f[l-1])*calc(tn/l);
			l=r+1;
		}
		// cout<<ans<<endl;
		cout<<(ans+(1ll<<32))%(1ll<<32)<<endl;
	}
	return 0;
}