Code(hdu5212)

Code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 877    Accepted Submission(s): 348

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

There are Multiple Cases.(At MOST 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

5

1 3 4 2 4

 Sample Output

64

思路：莫比乌斯反演；

F[x]表示sum(gcd(a,b))(x|gcd(a,b)),那么我们需要求f(x)（gcd(a,b)==x），那么根据莫比乌斯反演可得

f(x) = sum(u(y)F（y/x）)(x|y);然后答案就是sum（f(x)*（x*(x-1)））;所以我们先线性筛出莫比乌斯函数，然后再求每个数

的因数，复杂度（n*log（n））

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
typedef long long LL;
using namespace std;
bool prime[10005];
int ak[10005];
LL mul[10005];
LL cnt[10005];
vector<int>vec[10005];
int mod = 10007;
int main(void)
{
    int i,j;
    int cn = 0;
    mul[1] = 1;
    for(i = 2; i <= 10000; i++)
    {
        if(!prime[i])
        {
            ak[cn++] = i;
            mul[i] = -1;
        }
        for(j = 0; j < cn&&(LL)ak[j]*i<=10000; j++)
        {
            if(i%ak[j])
            {
                prime[i*ak[j]] = true;
                mul[i*ak[j]] = -mul[i];
            }
            else
            {
                prime[i*ak[j]] = true;
                mul[i*ak[j]] = 0;
                break;
            }
        }
    }
    int n;
    for(i = 1;i <= 10000;i++)
    {
        for(j = 1;j <= sqrt(i);j++)
        {
            if(i%j==0)
            {
                vec[i].push_back(j);
                if(i/j!=j)
                    vec[i].push_back(i/j);
            }
        }
    }
    while(scanf("%d",&n)!=EOF)
    {   int maxx = 0;
        memset(cnt,0,sizeof(cnt));
        int i,j;
        for(i = 0; i < n; i++)
            {scanf("%d",&ak[i]);maxx = max(maxx,ak[i]);}
        for(i = 0; i < n; i++)
        {
            for(j = 0;j < vec[ak[i]].size();j++)
            {   int x = vec[ak[i]][j];
                cnt[x]++;
            }
        }//printf("%lld
",cnt[1]);
        LL sum = 0;
        for(i = 1;i <= maxx;i++)
        {
            for(j = i;j <= maxx;j+=i)
            {
                sum = sum + mul[j/i]*(cnt[j]*cnt[j])*(i*(i-1)%mod)%mod;
                sum%=mod;
            }
        }
        printf("%lld
",(sum%mod+mod)%mod);
    }
    return 0;
}