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
Problem Description
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 .
Input
The input will contain a number 1 < x<= 65535 per
line representing the number of elements of the set.
Output
You have to print a line in the output for each entry
with the answer to the previous question.
Sample Input
11
9412
Sample Output
11
2*2*13*181
Author
eddy
Recommend
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。。
1 #include<stdio.h>
2 #include<string.h>
3 #define MAX 65543
4 bool flag[MAX] ;
5 int prime[MAX/2] ;
6 void get_prime( int &k )
7 {
8 memset(flag , true , sizeof (flag) ) ;
9 int i , j ;
10 for ( i = 2 ; i < MAX ; i ++ )
11 {
12 if ( flag[i] ) prime[k++] = i ;
13 for ( j = 0 ; j < k && i * prime[j] < MAX ; j ++ )
14 {
15 flag [i*prime[j]] = false ;
16 if ( i % prime[j] == 0 ) break ;
17 }
18 }
19 }
20
21 int main ()
22 {
23 int n , k = 0 ;
24 get_prime(k) ;
25 while ( scanf ( "%d" , &n ) != EOF )
26 {
27 int i ;
28 bool first = true ;
29 for ( i = 0 ; i < k ; i ++ )
30 {
31 while ( n % prime[i] == 0 )
32 {
33 if ( first )
34 {
35 printf ( "%d" , prime[i] ) ;
36 first = false ;
37 }
38 else printf ( "*%d" , prime[i] ) ;
39 n /= prime[i] ;
40 }
41 }
42 printf("\n");
43 }
44 return 0 ;
45 }
分类:
ACM
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 现代计算机视觉入门之:什么是视频
· 你所不知道的 C/C++ 宏知识
· 聊一聊 操作系统蓝屏 c0000102 的故障分析
· SQL Server 内存占用高分析
· .NET Core GC计划阶段(plan_phase)底层原理浅谈
· 我干了两个月的大项目,开源了!
· 千万级的大表,如何做性能调优?
· 盘点!HelloGitHub 年度热门开源项目
· Phi小模型开发教程:用C#开发本地部署AI聊天工具,只需CPU,不需要GPU,3G内存就可以运行,
· 你所不知道的 C/C++ 宏知识——基于《C/C++ 宏编程的艺术》