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
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+7;
int n,ans[N],q[N],top;
bool no[N],yes[N];
int main()
{
ios::sync_with_stdio(false);
for(int i=5; i<N; i+=4) {
if(no[i])continue;
q[++top]=i;
for(int j=5*i; j<N; j+=i*4) no[j]=1;
}
for(int i=1; i<=top; ++i)
for(int j=1; j<=i && q[i]*q[j]<N; ++j)
yes[q[i]*q[j]]=1;
for(int i=1; i<N; ++i)
ans[i]=ans[i-1]+yes[i];
while(555) {
cin>>n; if(!n) break;
cout<<n<<" "<<ans[n]<<"\n";
}
return 0;
}
原文地址：https://www.cnblogs.com/cytus/p/9572886.html