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  • 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

    21 
    85
    789
    0

    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
    ", h, HPrime[h]);
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/godweiyang/p/12203945.html
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