ACM--steps--2.1.4--素数筛选

Largest prime factor

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4866 Accepted Submission(s): 1450

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

12345

 

Sample Output

01213

 

Author

Wiskey

 

Source

HDU 2007-11 Programming Contest_WarmUp

 

Recommend

威士忌

问一个数的最大素因子是第几个素数。

算术基本定理：   任何大于1的正整数n可以唯一表示成有限个素数的乘积

值得注意的是，试了n遍，输入输出必须用scanf和printf不然，TIME LIMIT EXCEEDED

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int N=1000005;int wyx[N];int main(){    //进行类似于打表的计算。    memset(wyx,0,sizeof(wyx));    int i,j;    int k=0;    for(i=2;i<N;i++)    {        if(!(wyx[i]))//确定每个数字只会出现一次。        {//因为每个合数都可以分解成为两个其他数字的乘积。例如6=2*3；        k++;//k表示的是位置。        //接下来进行类似于打表的操作。        for(j=i;j<N;j+=i)        {            wyx[j]=k;        }        }    }    int n;    while(scanf("%d",&n)!=EOF)    {      printf("%d\n",wyx[n]);    }    return 0;}