抓其根本(一)（hdu2710 Max Factor 素数 最大公约数 最小公倍数.....）

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17     }8     return 0;          //是素数返回09 }

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110             }11         }12     }13 }

提供一种技巧、如果题目里面所有的计算都是素数之间的转化的话、可以如下。

 1 void prime() 2 { 3     int k=0; 4     for (int i=2; i<20050; i++)    5     { 6         if (hash[i]==0)             7         { 8             sushu[k++]=i;               //所有的素数都存在了sushu的数组里面,或者放在队列里面q.push(i); 9             for (int j=2; i*j<20050; j++) 10             {11                 hash[i*j]=1;              12             }13         }14     }15 }

 

 

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！！！