题解——GCD

给定正整数 $n$，求 $1\le x,y\le n$ 且 $gcd(x,y)$ 为素数的数对 $(x,y)$ 有多少对。

$n\le {10}^7$

题解

做法1

题意即为求$S=\sum _{\mathrm{质}\mathrm{数}p|n}\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=p]$

常规操作化简$S=\sum _{\mathrm{质}\mathrm{数}p|n}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{p}\rfloor }[gcd(i,j)=1]$

如果我们记$n^{\prime }=\lfloor \frac{n}{p}\rfloor ,f(i)=\sum _{a=1}^{n^{\prime }}\sum _{b=1}^{n^{\prime }}[gcd(a,b)=i],g(i)=\sum _{i|d}f(d)$

则展开$g(i)$有

$$g(i)=\sum _{i|d}\sum _{a=1}^{n^{\prime }}\sum _{b=1}^{n^{\prime }}[gcd(a,b)=d]$$

常规操作$g(i)=\sum _{a=1}^{n^{\prime }}\sum _{b=1}^{n^{\prime }}[i|gcd(a,b)]=\sum _{a=1}^{\lfloor \frac{n^{\prime }}{i}\rfloor }\sum _{b=1}^{\lfloor \frac{n^{\prime }}{i}\rfloor }[1|gcd(a,b)]={\lfloor \frac{n^{\prime }}{i}\rfloor }^2$

所以$S=\sum _{\mathrm{质}\mathrm{数}p|n}f(p)=\sum _{\mathrm{质}\mathrm{数}p|n}\sum _{d=1}^{dp\le n}\mu \left(d\right)g(p)=\sum _{\mathrm{质}\mathrm{数}p|n}\sum _{d=1}^{pd\le n}\mu (d){\lfloor \frac{n}{dp}\rfloor }^2$

接着有设$T=dp$,这里由于$p,d$具有对称性,交换求和次序有$S=\sum _{T=1}^n\lfloor \frac{n}{T}{\rfloor }^2\sum _{\mathrm{质}\mathrm{数}p|T}\mu \left(\frac{T}{p}\right)$

然后$\sum _{T=1}^n\lfloor \frac{n}{T}{\rfloor }^2$可以整除分块,而后面的可以在筛一遍质数之后$O(\sqrt{n})$求出,于是复杂度就是$O(n)$,实际上远远跑不满

做法2

事实上,因为数对具有对称性,我们可以将$S$稍稍变形得到

$S=\sum _{\mathrm{质}\mathrm{数}p|n}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{p}\rfloor }[gcd(i,j)=1]=\sum _{\mathrm{质}\mathrm{数}p|n}(2\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^i[gcd(i,j)=1]-1)$

这个$-1$是因为不能让$(p,p)$统计两次,根据$\phi$的定义,会发现原式等价于$\sum _{\mathrm{质}\mathrm{数}p|n}(2\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\phi (i)-1)$

预处理欧拉函数前缀和即可线性求解

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define N 10000005
#define int long long
int n, phi[N], prime[N], v[N], cnt, s[N];
int init() {
	phi[1] = 1;
	scanf("%lld", &n);
	for (int i = 2; i <= n; i++) {
		if (!v[i]) {
			v[i] = i;
			prime[++cnt] = i;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= cnt; j++) {
			if (prime[j] > v[i] || prime[j] * i > n)break;
			v[prime[j] * i] = prime[j];
			phi[prime[j] * i] = i % prime[j] ? phi[i] * (prime[j] - 1) : phi[i] * prime[j];
		}
	}
	for (int i = 1; i <= n; i++) {
		s[i] = s[i - 1] + phi[i];
	}
	return 0;
}
void solve() {
	int m = n, ans = 0;
	for (int i = 1; i <= cnt; i++) {
		ans += s[m / prime[i]] << 1;
		ans--;
	}
	printf("%lld", ans);
	return;
}
signed main() {
	init();
	solve();
	return 0;
}