poj3292(Semi-prime H-numbers)筛法

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

Sample Output

21 0
85 5
789 62

题意很好理解，用素数打表方法就行了。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <numeric>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

const int maxn = 1000010;

int HPrime[maxn];

void init()
{
    memset(HPrime, 0, sizeof(HPrime));
    for (int i = 5; i < maxn; i += 4)
        for (int j = i; j < maxn*1.0/i; j += 4)
            if (!HPrime[i] && !HPrime[j])
                HPrime[i*j] = 1;
            else
                HPrime[i*j] = -1;
    int cnt = 0;
    for (int i = 0; i < maxn; ++i)
        if (HPrime[i] == 1)
            HPrime[i] = ++cnt;
        else
            HPrime[i] = cnt;
}

int main()
{
    init();
    int h;
    while (scanf("%d", &h) == 1 && h)
        printf("%d %d\n", h, HPrime[h]);
    return 0;
}

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