（POJ3292）Semi-prime H-numbers <素数筛法的变形>

Semi-prime H-numbers

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

Waterloo Local Contest, 2006.9.30

题意：

H-number是4*n+1这样的数，如1，5，9，13… 。

H-primes是这样一个H-number：它只能唯一分解成1*它本身，而不能表示为其他两个H-number的乘积。

一个H-semi-prime是一个这样的H-number：它正好能表示成两个H-primes的乘积（除了1*它本身），这种表示法可以不唯一，但它不能表示为3个或者以上H-primes的乘积。

现在给出一个数n,要求区间[1,n]内有多少个H-semi-prime。

分析：

本题就是一个素数筛法的变形。原先我们求素数时是在1，2，3，。。。n中求解，不过现在变成了在1，5，9，。。。，4n+1里面求解。

所有照着素数筛法的算法写就可以了。

素数筛法：

int getPrime()
{
memset(prime,0,sizeof(prime));
for(int i=2;i<=maxn;i++)
{
if(!prime[i]) prime[++prime[0]] = i;
for(int j=1;j<=prime[0] && prime[j]<=maxn/i;j++)
{
prime[prime[j]*i] = 1;
if(i % prime[j] == 0) break;
}
}
return prime[0];
}

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int maxn = 1000010;
int ans[maxn],h;
int hprime[maxn];

void h_se_primes()
{
memset(hprime,0,sizeof(hprime));
for(int i=5;i<=maxn;i+=4)
{
for(int j=5;j<=i;j+=4)
{
if(i * j > maxn) break;
if(!hprime[i] && !hprime[j]) hprime[i*j] = 1;
else hprime[i*j] = -1;
}
}

//
memset(ans,0,sizeof(ans));
for(int i=1;i<=maxn;i++)
{
ans[i] = ans[i-1];
if(hprime[i]==1) ans[i]++;
}
}

int main()
{
h_se_primes();
while(scanf("%d",&h)!=EOF && h)
{
printf("%d %d\n",h,ans[h]);
}
return 0;
}