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uva 11426-欧拉函数

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Problem J GCD Extreme (II) 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: HereGCD(i,j) means the greatest common divisor of integeriand integer

Problem J
GCD Extreme (II)
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 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 Input                              Output for Sample Input

10

100

200000

0

 

67

13015

143295493160

 

代码:

#include<stdio.h>
#include<string.h>
#define MAXD 4000010
const int N = 4000000;
typedef long long LL;
int phi[MAXD];
LL a[MAXD];
void prep()
{
    memset(a, 0, sizeof(a));
    for(int i = 1; i <= N; i ++) phi[i] = i;
    for(int i = 2; i <= N; i ++)
    {
        if(phi[i] == i)
        {
            for(int j = i; j <= N; j += i)
                phi[j] = phi[j] / i * (i - 1);
        }
        for(int j = 1; j * i <= N; j ++)
            a[j * i] += j * phi[i];
    }
    for(int i = 1; i <= N; i ++) a[i] += a[i - 1];
}
int main()
{
    prep();
    int n;
    while(scanf("%d", &n), n) printf("%lld\n", a[n]);
    return 0;
}




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