Semi-prime H-numbers 筛选法的稍微变形

Problem 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

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筛选法&&二分查找

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 1 #include<iostream>
 2 #include<cmath>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cstring>
 6 using namespace std;
 7 //typedef long long LL;
 8 int t[1000002],HS[1000002];
 9 int HP[1000002];
10 int i,j,k,IP,v,IS;
11 int main()
12 {
13     memset(t,0,sizeof(t));
14     IP=0;
15    for(IP=0,i=5;i<1000;i+=4)
16     if(t[i]==0)
17      for(HP[IP++]=i,j=i*i;j<1000002;j+=i*4)
18        t[j]=1;
19    for(;i<1000002;i+=4)
20     if(t[i]==0)
21      HP[IP++]=i;//求素数
22    for(i=0;i<IP;i++)
23     for(v=1000002/HP[i],j=i;HP[j]<=v;j++)
24       t[HP[i]*HP[j]]=2;//求半素数
25    IS=1;
26    for(i=1;i<1000002;i+=4)
27     if(t[i]==2)
28       HS[IS++]=i;//记素数
29    //for(i=0;i<IS;i++)
30     //cout<<"HS[]:: "<<HS[i]<<endl;
31    int n,m,l,r;
32    for(;scanf("%d",&n)&&n>0;printf("%d %d\n",n,r))
33    {
34        for(l=105754,r=m=0;l-r>1;m=(l+r)>>1)//二分查找
35         if(HS[m]<=n)
36           r=m;
37         else
38            l=m;
39    }
40    return 0;
41 }

View Code

转载于:https://www.cnblogs.com/sdau--codeants/p/3389059.html