uva 11424 GCD Extreme 求∑∑gcd(i,j) (1<i<n,i<j<n)

Problem H

GCD Extreme

Input: Standard Input

Output: Standard Output

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); }

Input

The input file contains at most 20000 lines of inputs. Each line contains an integer N (1<N<200001). 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 Input                              Output for Sample Input

10 100              20000

67 13015 1153104356

Problemsetter: Shahriar Manzoor and Syed Monowar Hossain

Special Thanks: Shahriar Manzoor and Syed Monowar Hossain

#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;

int phi[1000100];

long long a[1000100];

void phi_table(int n )//筛选法求欧拉函数

{

 for(int i=2 ; i<= n; i++) phi[i] = 0;

 phi[1]= 1;

 for(int i =2; i<=n ;i ++)

  if(!phi[i])

  for(int j = i ; j<=n; j+=i)

  {

   if(!phi[j]) phi[j] = j;

   phi[j] = phi[j] /i*(i-1);

  }

}

void gcd_sum(int n)//求sigema gcd(i,n);(1<=i<=n-1);不包括n

{

    for(int i=2;i<n;i++) a[i]=phi[i];//

    double k=sqrt(0.5+n);

    for(int i=2;i<=k;i++)

    {

        for(int j=i,bit=1;j<n;j+=i,bit++)

        {

            if(i<bit) a[j]+=phi[bit]*i+phi[i]*bit;

            else if(i==bit) a[j]+=phi[i]*i;

        }

    }

}

void gcd_sum_again(int n)

{

    for(int i=1;i<n;i++) a[i]+=a[i-1];//二维的，要求和

}

int main()

{

    phi_table(1000100);

    gcd_sum(1000100);

    gcd_sum_again(1000100);

    int n;

    while(cin>>n&&n)

    {

        cout<<a[n]<<endl;

    }

    return 0;

}