poj3292——筛选法

poj3292——筛选法

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7772		Accepted: 3360

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 twoH-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
题意：不解释了，自己看吧
思路：先筛H素数，再筛Hsemi数，最后计数打表

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>



using namespace std;



const int maxn=1000100;

const int INF=(1<<28);



int h;

bool Hprime[maxn];

bool Hsemi[maxn];

int ans[maxn];



void play_list()

{

    memset(Hprime,1,sizeof(Hprime));

    for(int i=5;i<maxn;i+=4){///筛选H素数

        for(int j=5;j<maxn;j+=4){ ///把j的初值改成i就WA。。。因为下面的break使提前退出了？

            int mul=i*j;

            if(mul>=maxn||mul<=0) break;

            Hprime[mul]=0;

        }

    }

    memset(Hsemi,0,sizeof(Hsemi));

    for(int i=5;i<maxn;i+=4){///筛选Hsemi数

        for(int j=5;j<maxn;j+=4){ ///这里也是，把j的初值改成i就WA。。。

            int mul=i*j;

            if(mul>=maxn||mul<=0) break;

            if(Hprime[i]&&Hprime[j]) Hsemi[mul]=1;

        }

    }

    memset(ans,0,sizeof(ans));

    for(int i=1;i<maxn;i++){///打表计数

        if(Hsemi[i]) ans[i]=ans[i-1]+1;

        else ans[i]=ans[i-1];

    }

}



void debug()

{

    for(int i=100000;i<=100100;i++){

        cout<<i<<" "<<ans[i]<<endl;

    }

}



int main()

{

    play_list();

    //debug();

    while(cin>>h,h){

        cout<<h<<" "<<ans[h]<<endl;

    }

    return 0;

}

View Code