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 = ;

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