BZOJ2190 & 欧拉函数

题意:

　　求1-n内互质数对个数

SOL:

　　裸欧拉函数,还有莫比乌斯反演的加速什么的,挖个坑.

Code:

　　

/*==========================================================================
# Last modified: 2016-03-16 20:57
# Filename: 2190.cpp
# Description: 
==========================================================================*/
#define me AcrossTheSky 
#include <cstdio> 
#include <cmath> 
#include <ctime> 
#include <string> 
#include <cstring> 
#include <cstdlib> 
#include <iostream> 
#include <algorithm> 
  
#include <set> 
#include <map> 
#include <stack> 
#include <queue> 
#include <vector> 
 
#define lowbit(x) (x)&(-x) 
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) 
#define FORP(i,a,b) for(int i=(a);i<=(b);i++) 
#define FORM(i,a,b) for(int i=(a);i>=(b);i--) 
#define ls(a,b) (((a)+(b)) << 1) 
#define rs(a,b) (((a)+(b)) >> 1) 
#define getlc(a) ch[(a)][0] 
#define getrc(a) ch[(a)][1] 
 
#define maxn 100000 
#define maxm 100000 
#define pi 3.1415926535898 
#define _e 2.718281828459 
#define INF 1070000000 
using namespace std; 
typedef long long ll; 
typedef unsigned long long ull; 
 
template<class T> inline 
void read(T& num) { 
    bool start=false,neg=false; 
    char c; 
    num=0; 
    while((c=getchar())!=EOF) { 
        if(c=='-') start=neg=true; 
        else if(c>='0' && c<='9') { 
            start=true; 
            num=num*10+c-'0'; 
        } else if(start) break; 
    } 
    if(neg) num=-num; 
} 
/*==================split line==================*/ 
ll ans=3;
int n,p[maxn],pri[maxn],ma[maxn];

void prepare(){
	memset(pri,false,sizeof(pri));
	for (int i=2;i*i<=n;i++)
		if (!pri[i])
			for (int j=i*i;j<=n;j+=i)
				pri[j]=true,ma[j]=max(ma[j],i);
	FORP(i,2,n)
		if (!pri[i]) p[i]=i-1;
		else if (i/ma[i]%ma[i]==0) p[i]=p[i/ma[i]]*ma[i];
		else p[i]=p[i/ma[i]]*(ma[i]-1);
}

int main(){ 
	read(n); n--;
	prepare();
	FORP(i,1,n) ans+=p[i]*2;
	printf("%lld",ans);
}