素数判断:

一、根据素数定义，该数除了1和它本身以外不再有其他的因数。

详见代码。

1 int prime()
2 {
3     for (int i=2; i*i<=n; i++)
4     {
5         if (n%i==0)    //不是素数
6             return 1;  //返回1
7     }
8     return 0;          //是素数返回0
9 }

二、打表，将所有的素数一一列出，存在一个数组里。

详见代码。

 1 void prime()
 2 {
 3     for (int i=2; i<20050; i++)   //从2开始一个一个找
 4     {
 5         if (hash[i]==0)             //这一个判断可以减少很多重复的，节省很多时间
 6         {
 7             for (int j=2; i*j<20050; j++) //只要乘以i就一定不是素数
 8             {
 9                 hash[i*j]=1;               //不是素数标记为1
10             }
11         }
12     }
13 }

举个例子：hdu2710 Max Factor

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2710

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4168    Accepted Submission(s):
1366

Problem Description

To improve the organization of his farm, Farmer John
labels each of his N (1 <= N <= 5,000) cows with a distinct serial number
in the range 1..20,000. Unfortunately, he is unaware that the cows interpret
some serial numbers as better than others. In particular, a cow whose serial
number has the highest prime factor enjoys the highest social standing among all
the other cows.

(Recall that a prime number is just a number that has no
divisors except for 1 and itself. The number 7 is prime while the number 6,
being divisible by 2 and 3, is not).

Given a set of N (1 <= N <=
5,000) serial numbers in the range 1..20,000, determine the one that has the
largest prime factor.

Input

* Line 1: A single integer, N

* Lines 2..N+1:
The serial numbers to be tested, one per line

Output

* Line 1: The integer with the largest prime factor. If
there are more than one, output the one that appears earliest in the input
file.

Sample Input

4

36

38

40

42

Sample Output

38

题目大意：找到所给数的最大素因子，然后在比较这些素因子的大小，找到最大的，最后输出原有的那个数。

详见代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4
 5 using namespace std;
 6
 7 int hash[20050];
 8
 9 void prime()
10 {
11     for (int i=2; i<20050; i++)
12     {
13         if (hash[i]==0)
14         {
15             for (int j=1; i*j<20050; j++)
16             {
17                 hash[i*j]=i;//i表示的是最大的素因子
18             }
19         }
20     }
21     //return hash[n];
22 }
23
24 int main ()
25 {
26     int T;
27     memset(hash,0,sizeof(hash));
28     sushu();
29     while (~scanf("%d",&T))
30     {
31         int Max=0,x=1;
32         while (T--)
33         {
34             int n;
35             scanf("%d",&n);
36             if (hash[n]>Max)
37             {
38                 Max=hash[n];
39                 x=n;
40             }
41         }
42         printf ("%d\n",x);
43     }
44     return 0;
45 }

最大公约数（gcd）

详见代码。

1 int gcd(int a,int b)
2 {
3     return a%b?gcd(b,a%b):b;
4 } 

最小公倍数

求解最小公倍数，一般都要借助最大公约数。辗转相除求得最大公约数，再用两数之积除以此最大公约数，得最小公倍数。

注意基本！！！