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 10000000000LLinline 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\n",Ans);return 0;}