51Nod1584 加权约数和

这题其实就是反演一波就好了（那你还推了一下午+一晚上），不过第一次碰到(O(nlog n))预处理分块和式的方法……
不知为啥我跟唐教主的题解推的式子不太一样……（虽然本质上可能是相同的吧）
那就写一写好了，顺便骗点访问量（逃

[egin{align} onumber ext{Let}space A=&sum_{i=1}^nsum_{j=1}^i isigma_1(ij),B=sum_{i=1}^n isigma_1(i^2)\ onumber ext{Then}space Ans=&2A-B\ end{align}]

[egin{align} onumber A=&sum_{i=1}^nsum_{j=1}^i isigma_1(ij)\ onumber =&sum_{i=1}^nsum_{j=1}^i isum_{p|i}sum_{q|j}[(p,q)=1]frac{pj}q\ onumber =&sum_{d=1}^nmu(d)sum_{i=1}^nsum_{j=1}^i isum _{p|i}sum_{q|j}[d|(p,q)]frac{pj}q\ onumber =&sum_{d=1}^nmu(d)sum_{d|p}pleft(sum_{p|i}i ight)sum_{d|q}sum_{q|j}^i frac j q\ onumber =&sum_{d=1}^nmu(d)sum_{p=1}^{leftlfloorfrac n d ight floor}dpleft(sum_{p|i}^{leftlfloorfrac n d ight floor}di ight)sum_{q=1}^{leftlfloorfrac n d ight floor}sum_{q|j}^i frac{dj}{dq}\ onumber =&sum_{d=1}^nmu(d)sum_{p=1}^{leftlfloorfrac n d ight floor}dpleft(sum_{p|i}^{leftlfloorfrac n d ight floor}di ight)sum_{q=1}^isum_{q|j}^i frac j q\ onumber =&sum_{d=1}^nmu(d)d^2sum_{i=1}^{leftlfloorfrac n d ight floor}ileft(sum_{p|i}p ight)sum_{j=1}^isum_{q|j}frac j q\ onumber =&sum_{d=1}^nmu(d)d^2sum_{i=1}^{leftlfloorfrac n d ight floor}isigma_1(i)sum_{j=1}^isigma_1(j)\ onumber =&sum_{d=1}^nmu(d)d^2sum_{i=1}^{leftlfloorfrac n d ight floor}isigma_1(i)S_{sigma_1}(i) end{align}]

[egin{align} onumber B=&sum_{i=1}^n isigma_1(i^2)\ onumber =&sum_{i=1}^n isum_{p|i}sum_{q|j}[(p,q)=1]frac{pi}q\ onumber =&sum_{d=1}^nmu(d)sum_{i=1}^n isum_{p|i}sum_{q|j}[d|(p,q)]frac{pi}q\ onumber =&sum_{d=1}^nmu(d)sum_{d|i} isum_{d|p|i}sum_{d|q|j}frac{pi}q\ onumber =&sum_{d=1}^nmu(d)d^2sum_{i=1}^{leftlfloorfrac n d ight floor} isum_{p|i}sum_{q|j}pfrac i q\ onumber =&sum_{d=1}^nmu(d)d^2sum_{i=1}^{leftlfloorfrac n d ight floor} isigma_1^2(i)\ end{align}]

[Ans=2A-B\=sum_{d=1}^nmu(d)d^2sum_{i=1}^{leftlfloorfrac n d ight floor}isigma_1(i)left(2S_{sigma_1}(i)-sigma_1(i) ight) ]

到了这里就可以用(O(n))预处理+单次询问(O(sqrt n))的经典做法了，总复杂度(O(n+Tsqrt n))。
但是这样还是有点慢……常数优化到一定程度之后极限数据仍然要跑2s+（卡常技巧不行……逃），看来常数优化玩不了了，只能试试别的做法。
考虑枚举(d)和(k=leftlfloorfrac n d ight floor)，并考虑它们能对哪些(n)作出贡献。不难发现，如果(leftlfloorfrac n d ight floor=k)，那么一定有(nin[dk,d(k+1)))，再稍加观察就能得到每对((d,k))都会对(ge dk)的所有(n)作出贡献，因此枚举所有(d,k)并差分一下，最后求一遍前缀和即可。
这样就能做到预处理(O(nlog n))，询问(O(1))了，不用卡常也可以过。

#pragma GCC optimize("Ofast")
#include<stdio.h>
#define int unsigned
using namespace std;
template<class T>inline void readint(T &v){
	int x=0;
	char c=getchar();
	while(c<48)c=getchar();
	while(c>47){
		x=x*10+(c^48);
		c=getchar();
	}
	v=x;
}
template<class T>inline void writeint(T x){
	static int a[25];
	if(!x)putchar('0');
	else{
		int i=0;
		while(x){
			a[++i]=x%10;
			x/=10;
		}
		while(i)putchar(a[i--]^48);
	}
}
const int maxn=1000010,p=1000000007;
void get_table(int);
bool notp[maxn];
int prime[maxn],mu[maxn],sigma[maxn],ans[maxn],f[maxn];
signed main(){
	get_table(1000000);
	int T;
	readint(T);
	for(int k=1;k<=T;k++){
		int n;
		readint(n);
		putchar('C');
		putchar('a');
		putchar('s');
		putchar('e');
		putchar(' ');
		putchar('#');
		writeint(k);
		putchar(':');
		putchar(' ');
		writeint(ans[n]);
		putchar('
');
	}
	return 0;
}
void get_table(int n){
	mu[1]=sigma[1]=1;
	for(int i=2;i<=n;i++){
		if(!notp[i]){
			prime[++prime[0]]=i;
			mu[i]=-1;
			sigma[i]=i+1;
			f[i]=1;
		}
		for(int j=1;j<=prime[0]&&i*prime[j]<=n;j++){
			notp[i*prime[j]]=true;
			if(i%prime[j]){
				mu[i*prime[j]]=-mu[i];
				sigma[i*prime[j]]=(long long)sigma[i]*(prime[j]+1)%p;
				f[i*prime[j]]=i;
			}
			else{
				if(f[i]==1)sigma[i*prime[j]]=((long long)prime[j]*sigma[i]+1)%p;
				else sigma[i*prime[j]]=(long long)sigma[i/f[i]*prime[j]]*sigma[f[i]]%p;
				f[i*prime[j]]=f[i];
			}
		}
	}
	for(int i=2,sum=1;i<=n;i++){
		sum+=sigma[i];
		if(sum>=p)sum-=p;
		sigma[i]=(long long)i*sigma[i]%p*(2ll*sum-sigma[i]+p)%p;
	}
	for(int i=1;i<=n;i++){
		int t=((long long)i*i%p*(signed)mu[i]+p)%p;
		for(int j=1;i*j<=n;j++)ans[i*j]=(ans[i*j]+(long long)t*sigma[j])%p;
	}
	for(int i=2;i<=n;i++){
		ans[i]+=ans[i-1];
		if(ans[i]>=p)ans[i]%=p;
	}
}