hdu.5212.Code(莫比乌斯反演 && 线性筛）

时间：2015-05-04 08:35:42 阅读：100 评论：0 收藏：0 [点我收藏+]

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Code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 300 Accepted Submission(s): 124

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

Output

For each case:
Print an integer,denoting what the function returns.

Sample Input

5

1 3 4 2 4

Sample Output

64
Hint


gcd(x,y) means the greatest common divisor of x and y.

 1 #include<stdio.h>
 2 #include<string.h>
 3 int prime[10000 + 10] ;
 4 int a[10000 + 10] ;
 5 int mui[10000 + 10] ;
 6 bool vis[10000 + 10] ;
 7 
 8 int main ()
 9 {
10     memset (prime , 0 , sizeof(prime)) ;
11     memset (mui , 0 , sizeof(mui)) ;
12     memset (vis , 0 , sizeof(vis)) ;
13     for (int i = 1 ; i <= 10000 ; i ++) a[i] = i ;
14     for (int i = 2 ; i <= 10000 ; i ++) {
15         for (int j = i ; j <= 10000 ; j += i ) {
16             if (a[j] % i == 0 && ! vis[j] ) {
17                 int cnt = 0 ;
18                 while (a[j] % i == 0) {
19                     a[j] /= i ;
20                     cnt ++ ;
21                 }
22                 if (cnt > 1) { vis[j] = 1 ; mui[j] = 0 ;}
23                 else mui[j] ++ ;
24             }
25         }
26     }
27   /*  printf ("μ_source:\n") ;
28     for (int i = 2 ; i <= 5 ; i ++) printf ("ID %d: %d\n" , i , mui[i]) ; puts (""); */
29     mui[1] = 1 ;
30     for (int i = 2 ; i <= 10000 ; i++) {
31         if ( mui[i] ) mui[i] = (int) pow (-1 , mui[i]) ;
32     }
33 }