poj3292

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8677		Accepted: 3793

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

大致题意：

一个H-number是所有的模四余一的数。

如果一个H-number是H-primes 当且仅当它的因数只有1和它本身（除1外）。

一个H-number是H-semi-prime当且仅当它只由两个H-primes的乘积表示。

H-number剩下其他的数均为H-composite。

给你一个数h,问1到h有多少个H-semi-prime数。

解题思路：

感觉跟同余模扯不上关系。。。

筛法打表，再直接输出。。。水题。。。

ps:请用G++提交

#include<iostream>
using namespace std;
const int N=1000001;
int h,a[N+1];
int go(){
    for(int i=5;i<=N;i+=4){
        for(int j=5;j<=N;j+=4){
            int tmp=i*j;
            if(tmp>N) break;
            if(!a[i]&&!a[j])//i与j均为H-prime
                a[tmp]=1; //tmp为H-semi-primes
            else
                a[tmp]=-1;//tmp为H-composite
        }
    }
    int p=0; //H-prime计数器
    for(int i=1;i<=N;i++){
        if(a[i]==1) p++;
        a[i]=p; //从1到i有p个H-semi-primes
    }
}
int main(){
    go();
    while(cin>>h){
        if(!h) break;
        cout<<h<<' '<<a[h]<<endl;
    }
    return 0;
}