poj 3292(筛法+两两组合)

Semi-prime H-numbers
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6924   Accepted: 2959

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

Source

Waterloo Local Contest, 2006.9.30

题目比较简单,但却WA了好多次。
首先是对于H-prime的理解。首先H-prime必须是H-number(也就是模除4等于1),这个条件如果不满足的话,就无从谈起。
另外是与普通质数的对比。首先如果一个H-number同时也是质数,那它一定也是H-prime。比如5、13、17这样的数字。
反而,当一个H-number是合数, 那它也有可能是H-prime: 比如9。 这是因为他们无法拆分为两个H-number的乘积(3并不是H-number)。
因此判断H-number的方法就是: 能被4整除无法拆分为 2个H-number的乘积!

当找到了所有的H-prime,那么两两组合相乘,也就可以打表打出来MAXN以内的所有H-semi-primes来。这里不用担心MAXN*MAXN会tle,因为你设定好循环终止条件后,循环的效率会非常高,接近O(MAXN)。

提交记录:
1、WA 因为判断素数的时候条件写错,写成了遍历所有H-prime来判断能否找到两个H-prime的乘积等于num。其实这里并不要求 因子是H-prime,只要 是H-number就可以。
2、WA。主循环里 输入为0的时候需要跳出

AC代码:

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