POJ：3292-Semi-prime H-numbers（艾氏筛选法拓展）

Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 10466 Accepted: 4665

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

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解题心得：

1. 题意：
   • 如果一个数是4*n+1，那么这个数是H-number
   • 如果一个数是H-number且这个数是一个素数，那么这个数是H-prime
   • 如果一个数是H-number且这个数的因子仅仅有两个H-prime那么这个数是H-semi-prime（不包括1和他本身）
   • H-number剩下的数是H-composite
2. 其实就是一个艾氏筛选法的拓展，可以借鉴素数筛选。

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#include <algorithm>
#include <stdio.h>
#include <cstring>
using namespace std;
const int maxn = 1e6+100;
int prim[maxn];

void get_h_prim() {
    for(int i=5;i<maxn;i+=4)
        for(int j=5;j<maxn;j+=4) {
            long long temp = i*j;
            if(temp > maxn)
                break;
            if(prim[i] == prim[j] && prim[i] == 0)
                prim[temp] = 1;
            else
                prim[temp] = -1;
        }

    int cnt = 0;
    for(int i=0;i<maxn;i++) {
        if(prim[i] == 1)
            cnt++;
        prim[i] = cnt;
    }
}

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