Largest prime factor

Largest prime factor
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status
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

/*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 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
#include
using namespace std;
int su[1000010]={1,0};
void sushudabiao()
{
	int i,j,con=0;
	for(i=2;i<=1000000;i++)
	{
		if(su[i]==0){
			con++;
			su[i]=con;
			for(j=2*i;j<=1000000;j=j+i){
				su[j]=con;
			}
		}
	}
}

int main()
{
	sushudabiao();
	int n;
	while(~scanf("%d",&n))
	{
		printf("%d\n",su[n]);
	}
	return 0;
}