POJ 3292

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7059		Accepted: 3030

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

 

 

仿照素数的埃氏筛选法即可

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

#define maxn 1000005

bool H[maxn];
int ans[maxn],ele[maxn];
int len = 0;


void init() {

        for(int i = 1; i <= maxn - 5; i++) {
                H[i] = (i % 4 == 1);
        }

        for(int i = 5; i * i <= maxn - 5; i += 4) {
                if(!H[i]) continue;
                for(int j = i; j * i <= maxn - 5; j++) {
                        H[j * i] = 0;
                }
        }

        for(int i = 5; i <= maxn - 5; i += 4) {
                if(H[i]) {
                        ele[len++] = i;
                }
        }

        for(int i = 0; i < len && ele[i] * ele[i] <= maxn - 5; i++) {
                for(int j = i; j < len && ele[j] * ele[i] <= maxn - 5; j++) {
                        if(ele[i] * ele[j] % 4 == 1)
                        ans[ ele[i] * ele[j] ] = 1;
                }
        }

        for(int i = 0; i <=  maxn - 5; i++) {
                ans[i] += ans[i - 1];
        }
}

int main() {
       // freopen("sw.in","r",stdin);

        init();

        int x;
        while(~scanf("%d",&x) && x) {
                printf("%d %d
",x,ans[x]);
        }

        return 0;

}