POJ2480 Longge's problem

/*
POJ2480 Longge's problem
http://poj.org/problem?id=2480
数论 欧拉函数 线性筛 

求sigma_i=1^n{gcd(i,n)}
<=>sigma_d|n{d*phi(n/d)}
 *
 *
 *
 *
 */
#include <cstdio>
#include <algorithm>
#include <cmath>
#define new
#ifdef OLD
//250ms
using namespace std;
const long long Nmax=1000005LL;
long long n;
int is_prime[Nmax];
int prime[Nmax];
int cnt;
long long phi[Nmax];
long long oula(long long n)
{
    if(n==1LL)
        return 1LL;
    if(n<Nmax)
        return phi[n];
    long long ret=1LL,i;
    for(i=2LL;i*i<=n;i++)
    {
        if(n%i==0LL)
        {
            n/=i,ret*=i-1LL;
            while(n%i==0LL) n/=i,ret*=i;
        }
    }
    if(n>1LL) ret*=n-1LL;
    return ret;
}

void get()
{
    for(int i=2;i<Nmax;i++)
        is_prime[i]=1;
    for(long long i=2;i<Nmax;i++)
    {
        if(is_prime[i])
        {
            prime[++cnt]=i;
            phi[i]=(long long)(i-1);
        }
        for(int j=1;j<=cnt;j++)
        {
            if(i*prime[j]>=Nmax)
                break;
            is_prime[i*prime[j]]=0;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*(long long)prime[j];
                break;
            }
            else
                phi[i*prime[j]]=phi[i]*(long long)(prime[j]-1);
        }
    }
}

int main()
{
    // freopen("1.in","r",stdin);
    //scanf("%lld%lld",&a,&b);
    //printf("%lld
",gcd(a,b));
    get();
    while(scanf("%lld",&n)==1)
    {
        //n=read();
        //printf("%lld
",n);
        //if(n==0LL)
            //break;
        if(n<Nmax && is_prime[n])
        {
            printf("%lld
",2LL*n-1LL);
            continue;
        }
        long long ans=0LL;
        for(long long i=1LL;i*i<=n;i++)
        {
            if(n%i==0LL)
                ans+=i*oula(n/i)+n/i*oula(i);
            if(i*i==n)
                ans-=i*oula(n/i);
        }
        printf("%lld
",ans);
    }
    
    //for(long long  n=1;n<=100;n++)
    //{
        //long long ans=0LL;
        //for(long long i=1LL;i<=n;i++)
            //ans+=gcd(i,n);
        //printf("%d:%d
",n,ans);
        //long long m=n;
        ////for(long long i=1;prime[i])
    //}
    return 0;
}

#endif

#ifdef new
//32ms

using namespace std;
int main()
{
    long long n;
    while (scanf("%lld", &n) != EOF) 
    {
        long long i, sqr, p, a, ans;
        ans = n;
        sqr = floor(sqrt(n*1.0));
        for (i = 2; i <= sqr; i++) 
        {
            if (n%i == 0) 
            {
                a = 0;
                p = i;
                while (n%p == 0) 
                {
                    a++;
                    n /= p;
                }
                sqr = floor(sqrt(n*1.0));
                ans = ans + ans*a*(p-1)/p;
            }
        }
        if (n!=1) 
            ans = ans + ans*(n-1)/n;
        printf("%lld
", ans);
    }
    return 0;
}

#endif