【bzoj2818】Gcd 欧拉函数

AC通道：http://www.lydsy.com/JudgeOnline/problem.php?id=2818

【题解】 

用f[i]表示1~i中gcd(a,b)=1的数对(a,b)的对数，那么显然 f[i] = 1+2* sigma(phi[j])  1<j<=i

那么ans = sigma(f[N/p])  p为小于N的素数

原理很简单，即如果gcd(a,b)=1,那么gcd(a*p,b*p)=p

 

/*************  bzoj 2818  by chty  2016.11.3*************/#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>using namespace std;#define MAXN 10000010int n,cnt,prime[MAXN],check[MAXN],phi[MAXN];long long f[MAXN],ans;inline int read(){    int x=0,f=1;  char ch=getchar();    while(!isdigit(ch))  {if(ch=='-')  f=-1;  ch=getchar();}    while(isdigit(ch))  {x=x*10+ch-'0';  ch=getchar();}    return x*f;}void get(){    phi[1]=1;    for(int i=2;i<=n;i++)    {        if(!check[i])  {prime[++cnt]=i;  phi[i]=i-1;}        for(int j=1;j<=cnt&&prime[j]*i<=n;j++)        {            check[i*prime[j]]=1;            if(i%prime[j])  phi[i*prime[j]]=phi[i]*(prime[j]-1);            else {phi[i*prime[j]]=phi[i]*prime[j];  break;}        }    }}int main(){    freopen("cin.in","r",stdin);    freopen("cout.out","w",stdout);    n=read();    get();    f[1]=1;    for(int i=2;i<=n;i++)  f[i]=f[i-1]+2*phi[i];    for(int i=1;i<=cnt;i++)        if(prime[i]<n)  ans+=f[n/prime[i]];    printf("%lld\n",ans);    return 0;}