[51Nod

题面

求(sum_{i=1}^nsum_{j=1}^nvarphileft(gcd(varphi(i),varphi(j)) ight))
(nle 2e6)

题解

先初步推一推柿子。

[Ans=sum_{d=1}^nvarphi(d)sum_{i=1}^nsum_{j=1}^n[gcd(varphi(i),varphi(j))==d]$$用$s(k)$表示$n$以内$varphi(i)=k$的正整数$i$的个数。则 $$Ans=sum_{d=1}^nvarphi(d)sum_{i=1}^nsum_{j=1}^ns(i)cdot s(j)cdot[gcd(i,j)==d]\=sum_{d=1}^nvarphi(d)sum_{i=1}^{lfloorfrac nd floor}sum_{j=1}^{lfloorfrac nd floor}s(id)cdot s(jd)cdot[gcd(i,j)==1]$$令$g(d)=sum_{i=1}^{lfloorfrac nd floor}sum_{j=1}^{lfloorfrac nd floor}s(id)cdot s(jd)cdot[gcd(i,j)==1]$，发现不好求。 再令$f(d)=sum_{i=1}^{lfloorfrac nd floor}sum_{j=1}^{lfloorfrac nd floor}s(id)cdot s(jd)$，有： $$f(d)=sum_{k=1}^{lfloorfrac nd floor}g(kd)$$反演后得： $$g(d)=sum_{k=1}^{lfloorfrac nd floor}mu(k)f(kd)\ herefore Ans=sum_{i=1}^nvarphi(n)g(lfloorfrac nd floor)=sum_{d=1}^nvarphi(d)sum_{i=1}mu(i)f(id)]

那么预处理出来(f)，就能求(g)了。直接按照式子(O(nlog n))求。就完事了。

(O(Tnlog n))，有点慢还是能过。还有个小优化可以把(varphi)和(mu)枚举顺序交换一下，那么对于(mu=0)的时候就不用算（也优化不了多少）。

CODE

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2000005;
int n, cnt, p[N], phi[N], s[N], mu[N];
bool vis[N];
LL f[N];

void init(int N) {
	phi[1] = mu[1] = 1;
	for(int i = 2; i <= N; ++i) {
		if(!vis[i]) p[++cnt] = i, phi[i] = i-1, mu[i] = -1;
		for(int j = 1, k; j <= cnt && i * p[j] <= N; ++j) {
			vis[k = i * p[j]] = 1;
			if(i % p[j] == 0) {
				mu[k] = 0;
				phi[k] = phi[i] * p[j];
				break;
			}
			mu[k] = -mu[i];
			phi[k] = phi[i] * (p[j]-1);
		}
	}
}

int main () {
	init(2000000); int T, n;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		for(int i = 1; i <= n; ++i) ++s[phi[i]];
		for(int i = 1; i <= n; ++i) {
			f[i] = 0;
			for(int j = i; j <= n; j += i)
				f[i] += s[j];
			f[i] = f[i] * f[i];
		}
		LL ans = 0;
		for(int i = 1; i <= n; ++i) if(mu[i]) {
			LL sum = 0;
			for(int d = 1; i*d <= n; ++d)
				sum += phi[d] * f[i*d];
			ans += sum * mu[i];
		}
		printf("%lld
", ans);
		for(int i = 1; i <= n; ++i) --s[phi[i]];
	}
}