hdu.5212.Code(莫比乌斯反演 && 埃氏筛）

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 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

Hint

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

#include<stdio.h>
#include<string.h>
#include<math.h>
int F[10000 + 10] , f[10000 + 10];
int num[10000 + 10] ;
int prime[10000 + 10] ;
int a[10000 + 10] ;
int mui[10000 + 10] ;
bool vis[10000 + 10] ;
int b[10000 + 10] ;
int n ;
typedef long long ll ;
void work_miu ()
{
    memset (prime , 0 , sizeof(prime)) ;
    memset (mui , 0 , sizeof(mui)) ;
    memset (vis , 0 , sizeof(vis)) ;
    for (int i = 1 ; i <= 10000 ; i ++) a[i] = i ;
    for (int i = 2 ; i <= 10000 ; i ++) {
        for (int j = i ; j <= 10000 ; j += i ) {
            if (a[j] % i == 0 && ! vis[j] ) {
                int cnt = 0 ;
                while (a[j] % i == 0) {
                    a[j] /= i ;
                    cnt ++ ;
                }
                if (cnt > 1) { vis[j] = 1 ; mui[j] = 0 ;}
                else mui[j] ++ ;
            }
        }
    }
  /*  printf ("μ_source:
") ;
    for (int i = 2 ; i <= 5 ; i ++) printf ("ID %d: %d
" , i , mui[i]) ; puts (""); */
    mui[1] = 1 ;
    for (int i = 2 ; i <= 10000 ; i++) {
        if ( mui[i] ) mui[i] = (int) pow (-1 , mui[i]) ;
    }
}

void init() {
    mui[1] = 1;
    for (int i = 2; i <= 10000; ++ i) {
        int x = i;
        int cnt = 0;
        bool fuck = false;
        for (int j = 2; j * j <= i; ++ j) {
            if (x % j == 0) {
                x /= j;
                cnt ++;
                if (x % j == 0) {
                    fuck = true;
                    break;
                }
            }
        }
        if (x != 1) cnt ++;
        mui[i] = (cnt & 1) ? -1 : 1;
        if (fuck) mui[i] = 0;
    }
}

int main ()
{
   // freopen ("a.txt" , "r" , stdin ) ;
    work_miu () ;
    //init();
    while (~ scanf ("%d", &n)  )  {
        int x ;
        memset (F , 0 , sizeof(F)) ;
        memset (num , 0 , sizeof(num) ) ;
        memset (f , 0 , sizeof(f)) ;
        for (int i = 0 ; i < n ; i ++) {
                scanf ("%d" , &b[i]) ;
                num[ b[i] ] ++ ;
        }
        ll sum = 0 ;
        for (int i = 10000 ; i >= 1 ; i --) {
            int cnt = 0 ;
            for (int j = i ; j <= 10000 ; j += i) {
                cnt += num[j] ;
            }
            if (cnt > 0) {
                F[i] = (cnt * cnt - cnt) / 2 ;
                for (int j = i ; j <= 10000 ; j += i ) {
                    f[i] += mui[j / i] * F[j] ;
                }
           /*     if (f[i] != 0) {
                    printf ("ID %d : %d
" , i , f[i]) ;
                }*/
                sum += 1ll * f[i] * (i * i - i ) ;
            }
        } //puts ("") ;
      /*  puts ("F(X) :") ;
        for (int i = 1 ; i <= 5 ; i ++) printf ("ID %d: %d 
" , i , F[i]) ; puts ("") ;
        printf ("μ：
") ;
        for (int i = 1 ; i <= 5 ; i ++) printf ("ID %d: %d
" , i , mui[i]) ; puts ("") ;
        printf ("sum=%d
" , sum ) ; */
        sum *= 2 ;
        for (int i = 0 ; i < n ; i ++) {
            sum += b[i] * b[i] - b[i] ;
        }
        printf ("%I64d
" , sum % 10007) ;
    }
    return 0 ;
}

 复杂度这里，因为用的是埃帅，所以为O（nlog(log(n)) ) .

我从杰哥那里学到了一种和百度上不同的莫比乌斯反演写法(个人感觉不错）：

n为d的所有倍数。

则：

μ(1) = 1 ;

k = p1 * p2 * p3 ……*pr（k由r个不同的质数组成）则μ(k) = -1^k ;

其他情况，μ (k) = 0 ;

这道题F(x)指的是x的倍数的对数的个数有多少。

f(x) = 最大公约数为x的多数有多少。

比如：

F(1) = f(1) + f(2) + f(3) + f(4) = 7 + 2 + 0 + 1 = 10

得到F(x)是非常容易的可以统计x的倍数有多少个，比如说=cnt ；

那么此时的F(x) = (cnt * cnt - cnt) / 2 ;(稍微想想就能想通）

Then ， it's the time for 。。。。。