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;

}