Semi-prime H-numbers POJ - 3292

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

好题，以后还要重新再写一遍，筛法的思想

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

const int maxn=1000005;

int n;
int a[maxn],cnt[maxn];

void inite(){
    for(int i=5;i<maxn;i+=4){
        for(int j=5;i*j<maxn;j+=4){
            if(a[i]==0&&a[j]==0) a[i*j]=1;
            else a[i*j]=-1;
        }
    }
    int ans=0;
    for(int i=1;i<maxn;i++){
        if(a[i]==1) ans++;
        cnt[i]=ans;
    }
}

int main()
{   inite(); 
    while(cin>>n&&n) printf("%d %d\n",n,cnt[n]);
    return 0;
}