题目链接：http://poj.org/problem?id=3292
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
题意是1 5 9 13...这种4的n次方+1定义为H-numbers
H-numbers中只由1*自己这一种方式组成 即没有其他因子的 叫做H-prime
两个H-prime的乘积叫做H-semi-prime 还有一个要求是H-semi-prime只能由两个H-prime组成
代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1000001+5;
int sp[maxn],ip[maxn];
int main()
{
memset(ip,0,sizeof(ip));
for(int i=5;i<maxn;i+=4)
{
for(int j=5;i*j<maxn;j+=4)
{
if(ip[i]||ip[j]) ip[i*j]=-1;
else
ip[i*j]=1;
}
}
int num=0;
for(int i=1;i<maxn;i++)
{
if(ip[i]==1) num++;
sp[i]=num;
}
int h;
while(scanf("%d",&h)!=EOF&&h)
{
h=(h-1)/4*4+1;
printf("%d %d
",h,sp[h]);
}
return 0;
}