UVA 11426 GCD

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2421

gcd(x,4) f(4)=3,pos[4]=3=1*3;

gcd(2x,8) f[4]=3],pos[8]=6=2*3;

....................................;

pos[i*j]=pos[i]*j;

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll vis[4000006],pos[4000006],n;
void init()
{
    memset(pos,0,sizeof(pos));
    for(int i=1;i<=4000001;i++)
        vis[i]=i;
    for(int i=2;i<=4000001;i++)
    {
        if(vis[i]==i)
        {
            for(int j=1;j*i<4000006;j++)
            {
                vis[j*i]=vis[j*i]-vis[j*i]/i;
            }
        }
        for(int j=1;j*i<4000006;j++)
        {
            pos[j*i]+=j*vis[i];//倍数关系
        }
    }
}
int main()
{
    init();
    while(scanf("%lld",&n)&&n)
    {
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            ans+=pos[i];
        }
        printf("%lld
",ans);
    }
}