[51 Nod 1584] 加权约数和

题意

求$$large sum_{i=1}^Nsum_{j=1}^Nmax(i,j)cdotsigma_1(ij)$$
其中
$1le Nle10^6$
$1le Tle5cdot10^4$
$sigma_1(n)$表示$n$的约数和
### 题目分析
令$A=sum_{i=1}^nsum_{j=1}^iicdotsigma_1(ij),B=sum_{i=1}^nicdotsigma_1(i^2)$,则$Ans=2A-B$

且有
$$large
egin{aligned}
A&=sum_{i=1}^nicdotsum_{j=1}^isum_{x|i}sum_{y|j}xcdot j/y[(x,y)=1]\
&=sum_{i=1}^nicdotsum_{j=1}^isum_{x|i}sum_{y|j}xcdot j/ysum_{d|(x,y)}mu(d)\
&=sum_{d=1}^nmu(d)sum_{d|x}xsum_{x|i}isum_{d|y}sum_{y|j}^ifrac jy\
&=sum_{d=1}^nmu(d)sum_{x=1}^{lfloorfrac nd floor}dxsum_{x|i}^{lfloorfrac nd floor}disum_{y=1}^{lfloorfrac nd floor}sum_{y|j}^ifrac{dj}{dy}\
&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}isum_{x|i}xsum_{y=1}^{lfloorfrac nd floor}sum_{y|j}^iy\
&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}icdotsigma_1(i)sum_{j=1}^{i}sigma_1(j)\
&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}icdotsigma_1(i)cdot S_{sigma_1}(i)\end{aligned}$$
$$large
egin{aligned}
B&=sum_{i=1}^nisum_{x|i}sum_{y|i}xcdot i/y[(x,y)=1]\
&=sum_{i=1}^nicdotsum_{x|i}sum_{y|i}xcdot i/ysum_{d|(x,y)}mu(d)\
&=sum_{d=1}^nmu(d)sum_{d|i}isum_{d|x|i}sum_{d|y|i}xcdot frac iy\
&=sum_{d=1}^nmu(d)sum_{i=1}^{lfloorfrac nd floor}disum_{x|i}sum_{y|i}dxcdot frac {di}{dy}\
&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}isum_{x|i}xsum_{y|i}frac {i}{y}\
&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}isum_{x|i}xsum_{y|i}y\
&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}icdotsigma_1(i)^2\
end{aligned}$$
$$large
egin{aligned}
herefore Ans(n)&=2A-B\&=sum_{d=1}^nmu(d)d^2sum_{i=1}^{lfloorfrac nd floor}icdotsigma_1(i)(2cdot S_{sigma_1}(i)-sigma_1(i))
end{aligned}$$

发现$Ans(n)$可以递推：
$$Ans(n)=Ans(n-1)+sum_{d|n}mu(d)d^2{lfloorfrac nd floor}cdotsigma_1({lfloorfrac nd floor})(2cdot S_{sigma_1}({lfloorfrac nd floor})-sigma_1({lfloorfrac nd floor}))$$
那么就可以$O(nlog n)$预处理了。

然后每次$O(1)$回答就完了。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1000005;
const int mod = 1000000007;
int Cnt, Prime[N], mu[N];
LL ans[N], sd[N], a[N], d[N], f[N];
bool IsnotPrime[N];

void init() {
	d[1] = 1, a[1] = 1, mu[1] = 1;
	for(int i = 2; i < N; ++i) {
		if(!IsnotPrime[i])
			Prime[++Cnt] = i, d[i] = a[i] = 1+i, mu[i] = -1;
		for(int j = 1, k; j <= Cnt && i * Prime[j] < N; ++j) {
			IsnotPrime[k = i * Prime[j]] = 1;
			if(i % Prime[j] == 0) {
				mu[k] = 0;
				a[k] = a[i] * Prime[j] + 1;
				d[k] = d[i] / a[i] * a[k];
				break;
			}
			mu[k] = -mu[i];
			a[k] = 1 + Prime[j];
			d[k] = d[i] * a[k];
		}
	}
	for(int i = 1; i < N; ++i) {
		sd[i] = (sd[i-1] + (d[i]%=mod)) % mod;
		f[i] = i*d[i]%mod*(2*sd[i]-d[i]+mod)%mod;
	}
	for(int i = 1; i < N; ++i) if(mu[i])
		for(int j = i; j < N; j += i)
			ans[j] = (ans[j] + 1ll*mu[i]*i*i%mod*f[j/i]%mod + mod) % mod;
	for(int i = 1; i < N; ++i) ans[i] = (ans[i-1] + ans[i]) % mod;
}
int main () {
	init();
	int T, ks = 0, N; scanf("%d", &T);
	while(T--) scanf("%d", &N), printf("Case #%d: %lld
", ++ks, ans[N]);
}