Largest prime factor

Largest prime factor

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12064    Accepted Submission(s): 4282

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

 

方法就是 素数筛选法

就是从2开始，2为素数，则2的倍数4不为素数，3位素数，则3的倍数6不为素数。

2位第一个素数，则4为第一个非素数，所以倍数的序号是相同的。

用一个数组，把一开始都置为0，则 f [2]=1,f [4]=1;  f [3] =2,f [6] = 2;.........类推

#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;
}