POJ3292 Semi-prime H-numbers [数论，素数筛]

　　题目传送门

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10871		Accepted: 4881

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

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　　分析：

　　一道素数筛法的变式题。

　　把素数筛法改一下，预处理出所有答案，然后直接输出每个答案就行了。

　　Code：

//It is made by HolseLee on 2nd Sep 2018
//POJ3292
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
using namespace std;

const int N=1e6+;
int n,ans[N],q[N],top;
bool no[N],yes[N];

int main()
{
    ios::sync_with_stdio(false);
    for(int i=; i<N; i+=) {
        if(no[i])continue;
        q[++top]=i;
        for(int j=*i; j<N; j+=i*) no[j]=;
    }
    for(int i=; i<=top; ++i)
    for(int j=; j<=i && q[i]*q[j]<N; ++j)
    yes[q[i]*q[j]]=;
    for(int i=; i<N; ++i)
    ans[i]=ans[i-]+yes[i];

    while() {
        cin>>n; if(!n) break;
        cout<<n<<" "<<ans[n]<<"\n";
    }
    return ;
}