hdu 5212 Code

Code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 182

Problem Description

WLD likes playing with codes.One day he is writing a function.Howerver,his computer breaks down because the function is too powerful.He is very sad.Can you help him?

The function:

int calc 
 
 { 
 
    
 
   int res=0; 
 
    
 
   for(int i=1;i<=n;i++) 
 
      
 
     for(int j=1;j<=n;j++) 
 
      
 
     { 
 
        
 
       res+=gcd(a[i],a[j])*(gcd(a[i],a[j])-1); 
 
        
 
       res%=10007; 
 
      
 
     } 
 
    
 
   return res; 
 

 }

 

Input

10)

For each case:

The first line contains an integer N(1≤N≤10000).

The next line contains N integers a1,a2,...,aN(1≤ai≤10000).

 

Output

For each case:

Print an integer,denoting what the function returns.

 

Sample Input

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define MAX 10007
#define MOD 10007

using namespace std;

int f[MAX];
int n,a;
int cnt[MAX];

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
        memset ( cnt , 0 , sizeof ( cnt ) );
        for ( int i = 1 ; i <= n ; i++ )
        {
            scanf ( "%d" , &a );
            for ( int j = 1 ; j*j <= a ; j++ )
                if ( a%j == 0 )
                {
                    cnt[j]++;
                    if ( a/j != j )
                        cnt[a/j]++;
                }
        }
        int ans = 0;
        for ( int x = 10000 ; x > 0 ; x-- )
        {
            f[x] = cnt[x]*cnt[x]%MOD;
            for ( int i = x*2 ; i <= 10000 ; i += x )
                f[x] = ( f[x] - f[i] + MOD) %MOD;
            int p = x*(x-1)%MOD;
            ans = ( ans + p*f[x]%MOD )%MOD;
        }
        printf ( "%d\n" , ans );
    }
    return 0;
}