HDU 1164 Eddy's research I

Eddy's research I

Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 2486 Accepted Submission(s):
1527


[align=left]Problem Description[/align]

Eddy's interest is very extensive, recently he is
interested in prime number. Eddy discover the all number owned can be divided
into the multiply of prime number, but he can't write program, so Eddy has to
ask intelligent you to help him, he asks you to write a program which can do the
number to divided into the multiply of prime number factor .

[align=left]Input[/align]

The input will contain a number 1 < x<= 65535 per
line representing the number of elements of the set.

[align=left]Output[/align]

You have to print a line in the output for each entry
with the answer to the previous question.

[align=left]Sample Input[/align]

11
9412

[align=left]Sample Output[/align]

11
2*2*13*181

[align=left]Author[/align]

eddy

[align=left]Recommend[/align]

JGShining
素数筛选：对于素数筛选的几种方法，日后再做总结。这样不用每次回顾时想半天了。。

#include<stdio.h>
#include<string.h>
#define MAX 65543
bool flag[MAX] ;
int prime[MAX/2] ;
void get_prime( int &k )
{
memset(flag , true , sizeof (flag) ) ;
int i , j ;
for ( i = 2 ; i < MAX ; i ++ )
{
if ( flag[i] )  prime[k++] = i ;
for ( j = 0 ; j < k && i * prime[j] < MAX ; j ++ )
{
flag [i*prime[j]] = false ;
if ( i % prime[j] == 0 ) break ;
}
}
}

int main ()
{
int n , k = 0 ;
get_prime(k) ;
while ( scanf ( "%d" , &n ) != EOF )
{
int i ;
bool first = true ;
for ( i = 0 ; i < k ; i ++ )
{
while ( n % prime[i] == 0 )
{
if ( first )
{
printf ( "%d" , prime[i] ) ;
first = false ;
}
else printf ( "*%d" , prime[i] ) ;
n /= prime[i] ;
}
}
printf("\n");
}
return 0 ;
}