uva 11424 GCD Extreme 求∑∑gcd(i,j) (1

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

 

Input

The input file contains at most 20000 lines of inputs. Each line contains an integer N (1 

 

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 0

67 13015 1153104356

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Problemsetter: Shahriar Manzoor and Syed Monowar Hossain

Special Thanks: Shahriar Manzoor and Syed Monowar Hossain

#include
#include
#include
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    double k=sqrt(0.5+n);
    for(int i=2;i<=k;i++)
    {
        for(int j=i,bit=1;j        {
            if(i            else if(i==bit) a[j]+=phi[i]*i;
        }
    }
}
void gcd_sum_again(int n)
{
    for(int i=1;i}
int main()
{
    phi_table(1000100);
    gcd_sum(1000100);
    gcd_sum_again(1000100);
    int n;
    while(cin>>n&&n)
    {
        cout<    }
    return 0;
}