BZOJ 2818: Gcd

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题目

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2818: Gcd

Time Limit: 10 Sec  Memory Limit: 256 MB

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

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题解

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这题给的内存好极限，我调内存调了好久！【或者说我的方法太low了

求1<i,j<k的gcd(i,j)=1的数量，即为求phi[1~k]的和！

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代码

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/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;

#define LL long long
#define Inf 2147483647
#define InfL 10000000000LL

inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}

const int Maxn=10000000;

LL n;

int prime[Maxn/2+1];
bool valid[Maxn+1];
int primes;
inline void getPrime(){
	memset(valid,true,sizeof(valid));
	for (int i=2;i<=n;i++){
		if (valid[i])prime[++primes]=i;
		for (int j=1;j<=primes && prime[j]*i<=n;j++){
			valid[prime[j]*i]=false;
			if (i%prime[j]==0) break;
		}
	}
}

/*
LL miu[Maxn+10];
inline void getMiu(){
	for (int i=1;i<=Maxn;i++){
		int target=(i==1?1:0);
		int delta=target-miu[i];
		miu[i]=delta;
		for (int j=i+i;j<=Maxn;j+=i) miu[j]+=delta;
	}
}*/

LL phi[Maxn+1];
int minDiv[Maxn+1];
inline void getPhi(){
	for (int i=1;i<=prime[primes];i++) minDiv[i]=i;
	for (int i=2;i*i<=prime[primes];i++)
		if (minDiv[i]==i)
			for (int j=i*i;j<=prime[primes];j+=i)
				minDiv[j]=i;
	phi[1]=1;
	for (LL i=2;i<=prime[primes];i++){
		phi[i]=phi[i/minDiv[i]];
		if ((i/minDiv[i])%minDiv[i]==0){
			phi[i]*=minDiv[i];
		}else{
			phi[i]*=minDiv[i]-1;
		}
	}
}

LL Sum[Maxn+1];
int main(){
	scanf("%d",&n);
	getPrime();
	getPhi();
	for (int i=1;i<=Maxn;i++)Sum[i]=Sum[i-1]+phi[i];
	LL Ans=0;
	for (int i=1;prime[i]<=n&&i<=primes;i++){
		Ans+=Sum[n/prime[i]]*2-1;
	}
	printf("%lld
",Ans);
	return 0;
}