素数筛(hdu 2710)
素数
枚举因子法
bool judeg(int n)
{
if(n==1) return false;
for(int i=2; i*i<=n; i++)
{
if(n%i==0) return false;
}
return true;
}
更准确的说应该是判断某个数是不是素数,如果做素数筛用,复杂度会很高
普通素数筛
1.把1~n列出来
2.去掉不是特殊的1
3.从小到大,枚举每一个没有删掉的数字i
4.把i的2倍,3倍,4倍,…,删掉
5.剩下的没被删掉的都是素数
const int N=1000005;
bool prime[N];
void init()
{
memset(prime,false,sizeof(prime));
prime[1]=true;
for(int i=2; i<N; i++)
{
if(prime[i]==false)
{
for(int j=i+i; j<N; j=j+i) //素数的倍数全部筛去
{
prime[j]=true;
}
}
}
}
复杂度为O(nloglogn),缺点是有的合数被删除了多次,那么如何保证每个合数只被删除一次呢?
素数的线性筛
1.列出1~n
2.删除特殊的1
3.从2开始枚举数i,若没有被删掉,则加入素数表中
4.枚举素数表中的每一个素数j,并把i*j的结果从表里删去,直到枚举到j能整除i。
const int N=100000005;
bool prime[N];
int a[N],k=0; //a[N]存素数
void init()
{
memset(prime,false,sizeof(prime));
prime[1]=true;
for(int i=2; i<N; i++)
{
if(prime[i]==false)
{
a[k++]=i;
}
for(int j=0; j<k && a[j]*i<N; j++) //枚举素数表
{
prime[a[j]*i]=true;
if(i%a[j]==0) break; //若该数是某个素数的倍数,则退出循环
}
}
}
if(i%a[j]==0) break;
这是一个很简单有效的优化,原理基于“尽量只筛一次”。 在这里提出“最小质因子”的概念,然后规定每个合数都只被它的最小质因子筛出————这个定义很好地契
合了“尽可能只筛一次”的想法。
回到问题,当前如果出现"i%a[j]==0"的情况,由于枚举素数a[j]是从小到大的,那么可以说明a[j]是i的最小质因子。
此时如果不终止循环,那么a[j+1],a[j+2]...会与i相乘得到其他的合数然后筛掉,但是该合数的最小质因子不是
该素数,然而我们使用了这个素数将其筛掉,这样做会使得一个数被筛多次,降低了效率。
例如 20应该由2 * 10筛掉,而不是被4 * 5筛掉(素数2<素数5)。
即每一个素数都只被记录一次,每一个合数都只会被一个唯一的最小质因子标记一次因此是O(n)线性复杂度。
一道最大素数的水题 hdu 2710
Max Factor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10212 Accepted Submission(s): 3292
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.
(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
* 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
436384042
Sample Output
38
代码如下
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int prime[20005];
void init()
{
memset(prime,0,sizeof(prime));
prime[1]=1; //题中认为1也是素数
for(int i=2; i<20005; i++)
{
if(prime[i]==0)
{
for(int j=i; j<20005; j=j+i)
{
prime[j]=i; //数组里记录的是该数的最大质因子
}
}
}
}
int main()
{
int n;
init();
while(scanf("%d",&n)!=EOF)
{
int x=0,a,ans;
while(n--)
{
scanf("%d",&a);
if(prime[a]>x)
{
x=prime[a];
ans=a;
}
}
printf("%d\n",ans);
}
return 0;
}