POJ 3292 Semi-prime H-numbers（數）

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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 857890

Sample Output

21 085 5789 62

題意  所有可以表示為4*k+1（k>=0）的數都稱為“H數”  而在所有“H數”中只能被1和自身整除的H數稱為“H素數“   能表示成兩個”H素數“積的數又稱為”Semi-prime H數“

輸入n  求1到n之間有多少個”Semi-prime H數“;

方法  先打個H素數表  再用H素數表中的數依次相乘  得到的數都標記  再用一個數組儲存每個數以內的標記數   輸入n後直接讀數組就行了

#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;}