Semi-prime H-numbers(筛法)

Semi-prime H-numbers

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 8069 Accepted: 3479

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

类似素数筛

#include <set>
#include <map>
#include <list>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define PI cos(-1.0)
#define RR freopen("input.txt","r",stdin)
using namespace std;

typedef long long LL;

const int MAX = 1e6+100;

int vis[MAX];

int Dp[MAX];

int main()
{
    memset(vis,0,sizeof(vis));
    for(LL i=5;i<MAX;i+=4)//标记Semi-prime H-numbers
    {
        for(LL j=i;j<MAX;j+=4)
        {
            LL ans=i*j;
            if(ans>MAX)
            {
                break;
            }
            if(vis[i]==0&&vis[j]==0)
            {
                vis[ans]=1;
            }
            else
            {
                vis[ans]=-1;
            }
        }
    }
    Dp[0]=0;
    for(int i=1;i<MAX;i++)//记录从1-i之间的Semi-prime H-numbers个数
    {
        if(vis[i]==1)
        {
            Dp[i]=Dp[i-1]+1;
        }
        else
        {
            Dp[i]=Dp[i-1];
        }
    }
    int n;
    while(scanf("%d",&n)&&n)
    {
        printf("%d %d\n",n,Dp[n]);
    }
    return 0;
}