POJ 3292 Semi-prime H-numbers (筛法统计)

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

题意

一个H-number是所有的模四余一的数。

如果一个H-number是H-primes当且仅当它的因数只有1和它本身（除1外）。

一个H-number是H-semi-prime当且仅当它只由两个H-primes的乘积表示，其余数均为H-composite。

H-number剩下其他的数均为H-composite。

给你一个数h,问1到h有多少个H-semi-prime数。

思路

类似于素数筛法，把所有H-semi-prime打表筛出，然后统计区间个数之后直接输出。

AC 代码

#include <iostream>#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<string.h>using namespace std;#define MAXX 1100000int isprime[MAXX];int num[MAXX];void init() //筛法打表{    memset(isprime,0,sizeof(isprime));    for(int i=5; i<MAXX; i+=4)  //遍历所有的H-number        for(int j=5; j<MAXX; j+=4)        {            if(i*j>MAXX)break;  //超出最大范围            if(isprime[i]==0&&isprime[j]==0)    //如果i,j都是H-prime                isprime[i*j]=1;     //那么i*j是H-semi-primes            else isprime[i*j]=-1;   //否则i*j是H-composite        }    int count=0;    for(int i=1; i<MAXX; i++)    {        if(isprime[i]==1)count++;        num[i]=count;    }}int main(){    init();    int a;    while(~scanf("%d",&a)&&a)        printf("%d %d\n",a,num[a]);    return 0;}