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
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：
1、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。
2、H-semi-primes是由两个H-primes相乘得出的。
3、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。
解题思路：
打表
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
using namespace std;
typedef long long ll;
const int N=1000001;
ll h_prime[N+1],hashtable[N+1],ans[N+1];
int h;
void init(){ //找出h_prime数
for(ll i=5;i<=N;i+=4){
h_prime[i]=1;
}
for(ll i=5;i<=N;i+=4){
for(ll j=i*2;j<=N;j+=i){
h_prime[j]=0; //筛法
}
}
}
void maketable(){ //相乘即得到答案
for(ll i=5;i<=N;i+=4){
for(ll j=5;j<=N;j+=4){
if(i*j>N) break; //注意及时退出
if(h_prime[i]&&h_prime[j]){
hashtable[i*j]=1;
}
}
}
ll cnt=0; //数组统计
for(ll i=1;i<=N;i++){
if(hashtable[i]==1) cnt++;
ans[i]=cnt;
}
}
int main(){
init();
maketable();
while(cin>>h&&h){
cout<<h<<" "<<ans[h]<<endl; //注意输出格式
}
}