HDU2136_Largest prime factor【水题】【筛法求素数】

Largest prime factor

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7216    Accepted Submission(s): 2559

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

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。

思路：巧用素数打表法。用sum计算素数的序号，将素数连同他的倍

数一起置为它的素数序号，从小到大循环，这样数组里存放的序号就

是最大素数因子的序号了。

注意：初始化时令所有数为0，Prime[0] = Prime[1] = 1。

即Prime[i]为0是素数，Prime[i]为1为素数。改变之后Prime[i]为数i

的最大素数因子的序号。

#include<stdio.h>
#include<string.h>

int Prime[1000010],Primer[1000010];
void IsPrime()
{
    int sum = 0;
    for(int i = 2; i <= 1000000; ++i)
        if(Prime[i] == 0)
        {
            ++sum;
            Prime[i] = sum;
            for(int j = i+i; j <= 1000000; j+=i)
                Prime[j] = sum;
        }
}
int main()
{
    IsPrime();
    int N;
    while(~scanf("%d",&N))
    {
        printf("%d\n",Prime[N]);
    }
    return 0;
}