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：
H-number是所有的4K+1的数,K是整数。
H-primes 是当且仅当它的因数只有1和它本身（除1外）
H-semi-prime当且仅当它只由两个H-primes的乘积表示
H-number剩下其他的数均为H-composite。
如果一个H-number是H-primes 。一个H-number是H-semi-prime，给你一个数n,问1到n有多少个H-semi-prime数。
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double eps=1e-10;
const double pi= acos(-1.0);
const int MAXN=1e6+10;
int a[MAXN];
int b[MAXN];
void Init()
{
int i,j;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(i=5;i<=MAXN;i+=4){
for(j=5;j<=MAXN;j+=4){
if(i*j>MAXN) break;
if(a[i]==0&&a[j]==0)
a[i*j]=1;
else
a[i*j]=2;
}
}
int cnt=0;
for(i=1;i<=MAXN;i++){
if(a[i]==1)
cnt++;
b[i]=cnt;
}
}
int main()
{
int n;
Init();
while(~scanf("%d",&n)){
if(!n) break;
printf("%d %d\n",n,b[n]);
}
return 0;
}
POJ 3292-Semi-prime H-numbers(筛选法)
原文地址：http://blog.csdn.net/u013486414/article/details/46228431
