POJ3292

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 857890

Sample Output

21 085 5789 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\n",h,sp[h]);    }    return 0;}