HDU 2136 Largest prime factor（筛选法）-优快云博客

本文介绍了一种通过筛选法求解任意数最大素数因子位置的方法，详细解释了算法实现过程并提供了代码实例。

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Largest prime factor

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4026 Accepted Submission(s): 1415

Problem Description

Everybody knows any number can be combined by the prime number.
Now, your task is telling me what position of the largest prime factor.
The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc.
Specially, LPF(1) = 0.

Input

Each line will contain one integer n(0 < n < 1000000).

Output

Output the LPF(n).

Sample Input

1

2

3

4

5

Sample Output

0

1

2

1

3

Author

Wiskey

Source

HDU 2007-11 Programming Contest_WarmUp

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分析：用筛选法(分解素数技巧),求N的最大素数因子在素数表的位置

#include<iostream>
using namespace std;
int a[1000000];
int main()
{
    int k=1;//k初始化为1,位置   
     memset(a,0,sizeof(a));
    /*筛选位置*/  
    for(int i=2;i<1000000;i++)//要到N,而不是sqrt(N)
    {
        if(!a[i])
        {
          a[i]=k;//素数i的位置   
            for(int j=i+i;j<1000000;j+=i)
                a[j]=k;    //构造出j的暂时最大素数因子的位置 
            k++;
        }
    }
    int n;
    while(~scanf("%d",&n))
    {
        printf("%d\n",a[n]);
    }
    return 0;
}