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））

 1 #include <iostream>
 2 #include<stdio.h>
 3 #include<algorithm>
 4 #include<string.h>
 5 #include<queue>
 6 #include<set>
 7 #include<math.h>
 8 #include<vector>
 9 typedef long long LL;
10 using namespace std;
11 bool prime[10005];
12 int ak[10005];
13 LL mul[10005];
14 LL cnt[10005];
15 vector<int>vec[10005];
16 int mod = 10007;
17 int main(void)
18 {
19     int i,j;
20     int cn = 0;
21     mul[1] = 1;
22     for(i = 2; i <= 10000; i++)
23     {
24         if(!prime[i])
25         {
26             ak[cn++] = i;
27             mul[i] = -1;
28         }
29         for(j = 0; j < cn&&(LL)ak[j]*i<=10000; j++)
30         {
31             if(i%ak[j])
32             {
33                 prime[i*ak[j]] = true;
34                 mul[i*ak[j]] = -mul[i];
35             }
36             else
37             {
38                 prime[i*ak[j]] = true;
39                 mul[i*ak[j]] = 0;
40                 break;
41             }
42         }
43     }
44     int n;
45     for(i = 1;i <= 10000;i++)
46     {
47         for(j = 1;j <= sqrt(i);j++)
48         {
49             if(i%j==0)
50             {
51                 vec[i].push_back(j);
52                 if(i/j!=j)
53                     vec[i].push_back(i/j);
54             }
55         }
56     }
57     while(scanf("%d",&n)!=EOF)
58     {   int maxx = 0;
59         memset(cnt,0,sizeof(cnt));
60         int i,j;
61         for(i = 0; i < n; i++)
62             {scanf("%d",&ak[i]);maxx = max(maxx,ak[i]);}
63         for(i = 0; i < n; i++)
64         {
65             for(j = 0;j < vec[ak[i]].size();j++)
66             {   int x = vec[ak[i]][j];
67                 cnt[x]++;
68             }
69         }//printf("%lld\n",cnt[1]);
70         LL sum = 0;
71         for(i = 1;i <= maxx;i++)
72         {
73             for(j = i;j <= maxx;j+=i)
74             {
75                 sum = sum + mul[j/i]*(cnt[j]*cnt[j])*(i*(i-1)%mod)%mod;
76                 sum%=mod;
77             }
78         }
79         printf("%lld\n",(sum%mod+mod)%mod);
80     }
81     return 0;
82 }