Hdu.5212.code (Möbius inversion && linear sieve)

Code

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others) total submission (s): Accepted s Ubmission (s): 124

Problem Descriptionwld likes playing with codes. One day and 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;
}

Inputthere is multiple Cases. (At mostTen)
For each case:
The first line contains an integer n(1≤n≤10000) .
The next line containsNIntegers a1,a2,.. . ,aN(1≤ai≤10000) .

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

Sample Input51 3 4 2 4

Sample Output64 HintGCD (x, y) means the greatest common divisor of x and Y.

1#include <stdio.h>2#include <string.h>3 intprime[10000+Ten] ;4 inta[10000+Ten] ;5 intmui[10000+Ten] ;6 BOOLvis[10000+Ten] ;7 8 intMain ()9 {Tenmemset (Prime,0,sizeof(prime)); OneMemset (MUI,0,sizeof(MUI)); Amemset (Vis,0,sizeof(Vis)); -      for(inti =1; I <=10000; i + +) a[i] =i; -      for(inti =2; I <=10000; i + +) { the          for(intj = i; J <=10000; J + =i) { -             if(A[j]% i = =0&&!Vis[j]) { -                 intCNT =0 ; -                  while(A[j]% i = =0) { +A[J]/=i; -CNT + + ; +                 } A                 if(CNT >1) {Vis[j] =1; MUI[J] =0 ;} at                 ElseMUI[J] + + ; -             } -         } -     } -   /*printf ("μ_source:\n"); - for (int i = 2; I <= 5; i + +) printf ("ID%d:%d\n", I, Mui[i]); puts ("");*/ inmui[1] =1 ; -      for(inti =2; I <=10000; i++) { to         if(Mui[i]) mui[i] = (int) Pow (-1, Mui[i]); +     } -}

View Code

I learned from the Jiege a kind of and Baidu on the different Möbius inversion (personal feeling good):

n is all multiples of d.

The

μ (1) = 1;

K = P1 * P2 * P3 ... *pr (k consists of r different prime numbers) μ (k) = -1^k;

In other cases, μ (k) = 0;

The question F (x) refers to the number of logarithms in multiples of X.

f (x) = Greatest common divisor is the number of the majority of X.

Like what:

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

Getting f (x) is very easy to count the number of x multiples, such as =cnt;

then f (x) = (CNT * cnt-cnt)/2;(think about it a little bit)

Then, it's the time for .....

Hdu.5212.code (Möbius inversion && linear sieve)