2818: Gcd

2818: Gcd

Time Limit: 10 Sec   Memory Limit: 256 MB
Submit: 4048   Solved: 1784
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Description

给定整数N,求1<=x,y<=N且Gcd(x,y)为素数的
数对(x,y)有多少对.

Input

一个整数N

Output

如题

Sample Input

4

Sample Output

4

HINT

hint

对于样例(2,2),(2,4),(3,3),(4,2)


1<=N<=10^7

Source

湖北省队互测

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莫比乌斯反演一波
ans = ∑F[i] (i∈prime)
F[i]:以i为最大公约数的对数
F[I] =  ∑u[d/i]*[n/d]*[n/d]  (i|d)
ans = ∑[n/d]^2∑u[d/i]
对于第二个∑,维护一个前缀和就好
(强行用所有素数来筛,复杂度近似O(n)的,n以内的素数个数约为(n/logn))

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int maxn = 1E7 + 10;
typedef long long LL;

int tot,prime[maxn],mu[maxn],sum[maxn];
bool not_prime[maxn];
LL ans,n;

int main()
{
	#ifdef DMC
		freopen("DMC.txt","r",stdin);
	#endif
	
	cin >> n; mu[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (!not_prime[i]) {mu[i] = -1; prime[++tot] = i;}
		for (int j = 1; j <= tot; j++) {
			int Nex = i*prime[j];
			if (Nex > n) break;
			not_prime[Nex] = 1;
			mu[Nex] = mu[prime[j]]*mu[i];
			if (i % prime[j] == 0) {mu[Nex] = 0; break;}
		} 
	}
	for (int i = 1; i <= tot; i++)
		for (int j = 1; ; j++) {
			int Nex = j*prime[i];
			if (Nex > n) break;
			sum[Nex] += mu[Nex/prime[i]];
		}
	for (LL i = 1; i <= n; i++) ans += 1LL*sum[i]*(n/i)*(n/i);
	cout << ans;
	return 0;
}

讲道理这个式子单词计算复杂度可以降成根号来回答多组询问。。bulabulabula

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