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抓其根本(hdu2710 Max Factor 素数 最大公约数 最小公倍数.....)

时间:2015-02-28 18:08:32      阅读:206      评论:0      收藏:0      [点我收藏+]

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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!!!

抓其根本(hdu2710 Max Factor 素数 最大公约数 最小公倍数.....)

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原文地址:http://www.cnblogs.com/qq-star/p/4305983.html

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