题目大意：4n+1成为H数，H数中如果一个数字如果找不到另一个H数（1除外）作为约数，那么他就是H素数（1除外），否则是H合数；如果一个H数是两个H素数的乘积，那么他就是H合成数；请问n以内有多少个H合成数？
Description
This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.
An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,… are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.
As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.
For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.
Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it’s the product of three H-primes.
Input
Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.
Output
For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.
Sample Input
21 85 789 0
Sample Output
21 0 85 5 789 62
解题思路
首先，求出n以内的所有H素数；接着，任意两个H素数相乘，得到的数字k就是H合成数（%4==1、两个H素数的乘积），标记f[k]=2；最后遍历f数组，有多少个2就有多少个合成数。
因为是多组数据，可能有大量重复运算，我们可以用前缀和预处理前i个数字中，有多少个H合成数。
程序实现
枚举4n+1，暴力枚举他的约数，如果没有约数就是H素数，放入s数组中。s数组中的任意两个数（可以相同）相乘得到H合成数k，标记f[k]=2；计算f[k]的“前缀和”，f[k]=1表示+1，其他情况表示0。
筛选法求H素数：H合数可以分解成H素数的乘积，每找到一个H素数，存入s数组，并将其倍数标记为H合数。j/i表示倍数，该倍数也应该是H数，故j/i % 4也应该等于1。当然，这个条件不写也没问题，因为如果倍数不是H数，乘积也不可能是H数：i是H素数，那么i*b%4 = i%4 * (b%4) = b%4，标记多了外重循环也不会遍历到他。
不难发现，加1次i、2次i、3次i、4次i、5次i……余数分别是2、3、0、1、2……只有加4次i才是有效的，每次加4i可以减少循环次数。
