POJ3292Semi-prime H-numbers(筛选法)



Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8250 Accepted: 3569

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 857890

Sample Output

21 085 5789 62

题目大意：
定义一种数叫H-numbers，它是所有能除以四余一的数。
在H-numbers中分三种数：
H-primes: 这种数只能被1和它本身整除，不能被其他的H-number整除，例如9是一个H-number，能被1，3，9整除，但3不是H-number，所以他是H-primes。
H-semi-primes是由两个H-primes相乘得出的。
剩下的是H-composite。
问给一个数，求1到这个数之间有多少个H-semi-primes。
筛选法，好题
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int MAX = 1000000 + 5;int a[MAX],cnt[MAX];int main(){    memset(a,0,sizeof(a));    memset(cnt,0,sizeof(cnt));    for(int i = 5; i < MAX; i+=4)    {        for(int j = 5; i * j < MAX; j+=4)        {            if(a[i] == 0 && a[j] == 0)//都没筛选掉，所以他们的乘积可以            {                a[i * j] = 1;            }            else                a[i * j] = -1;        }    }    int k = 0;    for(int i = 1; i < MAX; i ++)    {        if(a[i] == 1)            k++;        cnt[i] = k;    }    int n;    while(scanf("%d", &n) != EOF && n)    {        printf("%d %d\n",n,cnt[n]);    }    return 0;}