POJ 3292 Semi-prime H-numbers【筛选素数变形】
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8562 | Accepted: 3738 |
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
Source
题意：
题目给出的是几种定义，把所有的4*n+1 形式的自然数定义为H-number，在此基础上定义如下：
H-primes 有以下性质的数：
1,，不为1；
2，除了1 和自身，不能写成其他两个数的乘积；
H-semi-prime:
能写成两个（可以相同）H-primes 的乘积
然后给出一个一个H-number，问这个区间内有多少个 H-semi-prime
题解：
首先想到的就是基于自然数的素数的筛选的方法，先标记上H-primes，然后再进行H-semi-prime 的筛选
一开始想的太天真了，想把两个 步骤放在一起，结果发现，根本行不通，筛选素数的那种方法，在没有筛选出所有的素数的时候，是不能使用进行对素数的判定的，因为还没筛选出来，其实筛选到n 的时候，只有1-n*n 的区域是完全确定的，其他区间都不能保证，因此只能选择用两次筛选进行控制了...也许是个人对素数的理解还不够，所以找不到更优化的方法，希望有路过的大神指点，谢谢！
步骤如下（打表）：
1，模拟素数的筛选筛选出题目所定义的H-primes
2，进行遍历打表，筛选出所有的定义的H-semi-prime，并用数组实现循环累积
/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=1000001+5;
ll prime[maxn],num[maxn];
int main()
{
for(ll i=5;i<maxn;i+=4)
{
if(!prime[i])
{
for(ll j=i*i;j<maxn;j+=4*i)
{
prime[j]=1;
}
}
}
for(ll i=5;i<maxn;i+=4)
{
num[i]+=num[i-4];
if(!prime[i])
{
for(ll j=i*i;j<maxn;j+=4*i)
{
if(!prime[j/i])
{
num[j]=1;
}
}
}
}
ll n;
while(scanf("%lld",&n),n)
{
printf("%lld %lld\n",n,num[n]);
}
return 0;
}