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。

int prime()
{
    for (int i=2; i*i<=n; i++)
    {
        if (n%i==0)    //不是素数
            return 1;  //返回1
    }
    return 0;          //是素数返回0
}

二、打表，将所有的素数一一列出，存在一个数组里。

详见代码。

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

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。

void prime()
{
    int k=0;
    for (int i=2; i<20050; i++)   
    {
        if (hash[i]==0)            
        {
            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);
            for (int j=2; i*j<20050; j++) 
            {
                hash[i*j]=1;              
            }
        }
    }
}

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。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int hash[20050];

void prime()
{
    for (int i=2; i<20050; i++)
    {
        if (hash[i]==0)
        {
            for (int j=1; i*j<20050; j++)
            {
                hash[i*j]=i;//i表示的是最大的素因子
            }
        }
    }
    //return hash[n];
}

int main ()
{
    int T;
    memset(hash,0,sizeof(hash));
    sushu();
    while (~scanf("%d",&T))
    {
        int Max=0,x=1;
        while (T--)
        {
            int n;
            scanf("%d",&n);
            if (hash[n]>Max)
            {
                Max=hash[n];
                x=n;
            }
        }
        printf ("%d
",x);
    }
    return 0;
}

最大公约数（gcd）

详见代码。

int gcd(int a,int b)  
{  
    return a%b?gcd(b,a%b):b;  
} 

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！！！