uva 11417 - GCD

GCD
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<501). The meaning of N is given in the problem statement. Input is terminated by a line containing a single zero.  This zero should not be processed.

 

Output

For each line of input produce one line of output. This line contains the value of G for corresponding N.

 

Sample Input                              Output for Sample Input

10 100 500 0

67 13015 442011

---

Problemsetter: Shahriar Manzoor

Special Thanks: Syed Monowar Hossain

 

 

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
const int maxn = 503;

int G[maxn];
int opl[maxn];
void init()
{
    int i,j;
    memset(G,0,sizeof(G));
    for(i=2;i<maxn;i++) opl[i] = i;
    for(i=2;i<maxn;i++)
    {
        if(opl[i]==i)
        {
            for(j=i;j<maxn;j=j+i)
                opl[j] = opl[j]/i*(i-1);
        }
        for(j=1;i*j<maxn;j++)/**opl [ i  ] 此时已经算出**/
            G[j*i] = G[j*i] + opl[i]*j;
    }
    for(i=3;i<maxn;i++)
        G[i] +=G[i-1];
}
int main()
{
    int n;
    init();
    while(scanf("%d",&n)>0)
    {
        if(n==0)break;
        printf("%d\n",G[n]);
    }
    return 0;
}