Semi-prime H-numbers
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
题意 所有可以表示为4*k+1（k>=0）的数都称为“H数” 而在所有“H数”中只能被1和自身整除的H数称为“H素数“ 能表示成两个”H素数“积的数又称为”Semi-prime H数“
输入n 求1到n之间有多少个”Semi-prime H数“;
方法 先打个H素数表 再用H素数表中的数依次相乘 得到的数都标记 再用一个数组保存每个数以内的标记数 输入n后直接读数组就行了
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=1000001;
int vis[N],hp[N],ans[N],n;
int main()
{
int num=0,m=sqrt(N+0.5);
for(int i=5;i<=m;i+=4)
{
if(vis[i]==0)
for(int j=i*i;j<=N;j+=i)
vis[j]=1;
}
for(int i=5;i<N;i+=4)
if(!vis[i]) hp[++num]=i;
memset(vis,0,sizeof(vis));
for(int i=1;hp[i]*hp[i]<=N;++i)
for(int j=i;hp[i]*hp[j]<=N;++j)
++vis[hp[i]*hp[j]];
num=0;
for(int i=1;i<N;++i)
{
if(vis[i]>=1) ++num;
ans[i]=num;
}
while(scanf("%d",&n),n)
printf("%d %d\n",n,ans[n]);
return 0;
}
POJ 3292 Semi-prime H-numbers（数）,布布扣,bubuko.com