[uva 11426] GCD

A - GCD - Extreme (II)

 

Description

Given the value of N, you will have to find the value of G. The definition of G is given below:

Here GCD(i,j) means the greatest common divisor of integer i and integer j.

For those who have trouble understanding summation notation, the meaning of G is given in the following code: 

G=0;

for(i=1;i<N;i++)

for(j=i+1;j<=N;j++)

{

    G+=gcd(i,j);

}

/*Here gcd() is a function that finds the greatest common divisor of the two input numbers*/

Input

The input file contains at most 100 lines of inputs. Each line contains an integer N (1<N<4000001). The meaning of N is given in the problem statement. Input is terminated by a line containing a single zero. 

Output

For each line of input produce one line of output. This line contains the value of G for the corresponding N. The value of G will fit in a 64-bit signed integer.

Sample

10            67

100          13015

200000     153295493160

0

枚举最大公约数、利用欧拉函数

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define ll long long
#define N 4000000

int tot;
int prime[N+10];
bool isprime[N+10];
int phi[N+10];
void prime_pri()
{
    tot=0;
    phi[1]=1;
    memset(isprime,1,sizeof(isprime));
    isprime[0]=isprime[1]=false;
    for(int i=2;i<=N;i++)
    {
        if(isprime[i])
        {
            prime[tot++]=i;
            phi[i]=i-1;
        }
        for(int j=0;j<tot;j++)
        {
            if((ll)i*prime[j]>N) break;
            isprime[i*prime[j]]=0;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
            {
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
}
int n;
ll tmp[N+10];
ll ans[N+10];
int main()
{
    prime_pri();
    for(int i=1;i<=N;i++) //枚举gcd
    {
        for(int j=i+i;j<=N;j+=i)
        {
            tmp[j]+=(ll)phi[j/i]*i;
        }
    }
    for(int i=2;i<=N;i++) ans[i]=ans[i-1]+tmp[i];
    while(scanf("%d",&n),n)
    {
        printf("%lld
",ans[n]);
    }
    return 0;
}