POJ3292 筛法

Semi-prime H-numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8338 Accepted: 3621

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：
H-number：模4等于1的数
H-primes：是H-number且是素数
H-semi-prime：当且仅当只由两个H-primes表示
H-composite：除上述三种数以外的其他数
给一个数n，求从1到n总共有多少个H-semi-prime
题解：
类似埃式筛法，先将素数全部筛出来。然后素数间两两相乘。统计个数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <queue>#include <vector>#define f(i,a,b) for(i = a;i<=b;i++)#define fi(i,a,b) for(i = a;i>=b;i--)using namespace std;#define SIZE 1000100int p[SIZE];void init(){    memset(p,0,sizeof(p));    int i,j;    for(i = 5;i<SIZE;i+=4){        if(p[i]==0){            int k = 5;            for(j = i*2;j<=SIZE;j+=i)                p[j] = -1;        }    }    for(i = 5;i<SIZE;i+=4)        for(j = 5;j<SIZE&&i*j<SIZE;j+=4)            if(p[i]==0&&p[j]==0){                p[i*j] = 1;            }    int sum = 0;    f(i,1,SIZE){        if(p[i]==1)            sum++;        p[i] = sum;    }}int main(){//    freopen("data.out","w",stdout);    init();    int n;    while(~scanf("%d",&n)&&n)        printf("%d %d\n",n,p[n]);    return 0;}