| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9182 | Accepted: 4065 |
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
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，其余数均为H-composite。
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。
思路
类似于素数筛法，把所有H-semi-prime打表筛出，然后统计区间个数之后直接输出。
AC 代码
#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define MAXX 1100000
int isprime[MAXX];
int num[MAXX];
void init() //筛法打表
{
memset(isprime,0,sizeof(isprime));
for(int i=5; i<MAXX; i+=4) //遍历所有的H-number
for(int j=5; j<MAXX; j+=4)
{
if(i*j>MAXX)break; //超出最大范围
if(isprime[i]==0&&isprime[j]==0) //如果i,j都是H-prime
isprime[i*j]=1; //那么i*j是H-semi-primes
else isprime[i*j]=-1; //否则i*j是H-composite
}
int count=0;
for(int i=1; i<MAXX; i++)
{
if(isprime[i]==1)count++;
num[i]=count;
}
}
int main()
{
init();
int a;
while(~scanf("%d",&a)&&a)
printf("%d %d\n",a,num[a]);
return 0;
}
