Poj 3292 semi-prime H-Numbers

Semi-prime H-Numbers Time Limit: 1000 msmemory limit: 65536 K Total submissions: 7372 accepted: 3158 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 shocould 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 random Sive, 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

Question: it is hard to understand the meaning of the question !! It is to find the number from 1 to the input. It can only be split into two numbers that multiply by 4n + 1 !!

However, this question is very meaningful. It makes me deeply understand the power of table playing. This is another problem that I encountered after Queen n !!

The AC code is as follows:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define M 1000001using namespace std;int aa[M+1];int bb[M+1];int main(){    int i,j;    int n;    int t=0,tj=0;    memset(aa,0,sizeof aa);    for(i=5;i<=M;i+=4)        for(j=5;j<=M;j+=4)        {            if(i*j>M)                break;            if(aa[i]==0&&aa[j]==0)                aa[i*j]=1;            else aa[i*j]=-1;        }    int ans=0;    for(i=1;i<=M;i++)    {        if(aa[i]==1)        {            ans++;        }        bb[i]=ans;    }    while(~scanf("%d",&n)&&n)    {        printf("%d %d\n",n,bb[n]);    }    return 0;}

Poj 3292 semi-prime H-Numbers