HDU 2136 Largest prime factor （筛选法求素数）

Largest prime factor

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7297    Accepted Submission(s): 2589

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

 

题目大意：每个素数在素数表中都有一个序号，设1的序号为0，则2

的序号为1，3的序号为2，5的序号为3，以此类推。现在要求输出所

给定的数n的最大质因子的序号，0<n<1000000。

#include<iostream>#include<stdio.h>using namespace std;int pi;bool flag[1000010];int in;int n;int a[1000010];int getprime(){    for(int i=2;i<1000010;++i)    {        if(!flag[i])        {            in++;            for(int j=i;j<1000010;j+=i)            {                flag[j]=true;                a[j]=in;//记录序号            }        }    }}int main(int argc, char *argv[]){    //freopen("2136.in","r",stdin);    getprime();    while(scanf("%d",&n)!=EOF)    {        printf("%d\n",a[n]);    }    return 0;}