[poj3292]Semi-prime H-numbers（数论，素数）

Semi-prime H-numbers

Time Limit: 1000MS
Memory Limit: 65536K

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

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解题思路

题意理解了好一会儿……不过还是一道挺有趣的题
我们发现，与我们通常所讨论的正整数域相比，这道题只不过是把数域缩小为所有$4n+1$形式的正整数构成的集合上而已，素数、合数在新的数域上的定义并没有改变，那么我们在正整数域是怎么做的，在这道题里如法炮制就行了。
发现，在线性筛筛出素数的同时，可以顺便把所谓的semi-prime处理出来（见代码），最后前缀和优化一下以便$O(1)$回答查询

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Code

#include<cstdio>

using namespace std;

const int N = 1e6+5;
int h;
int pList[N], pNum, cnt[N];
bool notP[N], isHsp[N];
void Euler(){
    notP[1] = 1;
    for(int i = 1; i <= 1e6+1; i += 4){
        if(!notP[i])    pList[++pNum] = i;
        for(int j = 1; j <= pNum; j++){
            if(pList[j] * i > 1e6+1)    break;
            notP[pList[j]*i] = 1;
            if(!notP[i])    isHsp[pList[j]*i] = 1;//处理出semi-prime
            if(i % pList[j] == 0)   break;
        }
    }
}

int main(){
    Euler();
    for(int i = 1; i <= 1e6+1; i++)
        cnt[i] = cnt[i-1] + isHsp[i];
    while(scanf("%d", &h) && h)
        printf("%d %d\n", h, cnt[h]);
    return 0;
}