POJ 3292 Semi-prime H-numbers(类素数筛法)
2016-08-23 16:15
316 查看
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 areH-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
Sample Output
H-numbers是一类4n+1的数(在这个问题里,只有这类数字)
H-primes是一类因子只有1和它本身的H-numbers(类似于平常我们见到的质数)
H-semi-primes是一类数是两个H-primes的乘积
打个表,然后输出就可以
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int H_number[1000011];
void fun()
{
int i,j;
memset(H_number,0,sizeof(H_number));
for(i=5;i<=1000001;i+=4)
{
for(j=5;j<=1000001;j+=4)
{
int t = i*j;
if(t > 1000001)
break;
if(H_number[i] == 0 && H_number[j] == 0)
H_number[t] = 1;
else
H_number[t] = -1;
}
}
int cnt = 0;
for(i=1;i<=1000001;i++)
{
if(H_number[i] == 1)
{
cout << i << endl;
cnt++;
}
H_number[i] = cnt;
}
}
int main(void)
{
int n;
fun();
while(scanf("%d",&n)&&n)
{
printf("%d %d\n",n,H_number
);
}
return 0;
}
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 areH-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
H-numbers是一类4n+1的数(在这个问题里,只有这类数字)
H-primes是一类因子只有1和它本身的H-numbers(类似于平常我们见到的质数)
H-semi-primes是一类数是两个H-primes的乘积
打个表,然后输出就可以
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int H_number[1000011];
void fun()
{
int i,j;
memset(H_number,0,sizeof(H_number));
for(i=5;i<=1000001;i+=4)
{
for(j=5;j<=1000001;j+=4)
{
int t = i*j;
if(t > 1000001)
break;
if(H_number[i] == 0 && H_number[j] == 0)
H_number[t] = 1;
else
H_number[t] = -1;
}
}
int cnt = 0;
for(i=1;i<=1000001;i++)
{
if(H_number[i] == 1)
{
cout << i << endl;
cnt++;
}
H_number[i] = cnt;
}
}
int main(void)
{
int n;
fun();
while(scanf("%d",&n)&&n)
{
printf("%d %d\n",n,H_number
);
}
return 0;
}
相关文章推荐
- 素数判定 & 素数筛法 & poj_3292_Semi-prime H-numbers
- POJ - 3292 Semi-prime H-numbers(素数筛法)
- POJ 3292 Semi-prime H-numbers (类似素数筛)
- POJ 3292:Semi-prime H-numbers 筛选数
- POJ-3292-Semi-prime H-numbers
- Semi-prime H-numbers(POJ--3292
- poj3292——Semi-prime H-numbers(数论)
- POJ 3292, Semi-prime H-numbers
- POJ 3292.Semi-prime H-numbers
- poj 3292 Semi-prime H-numbers
- POJ 3292 Semi-prime H-numbers (筛法统计)
- POJ 3292 Semi-prime H-numbers
- poj 3292 Semi-prime H-numbers(筛法~)
- POJ-3292-Semi-prime H-numbers
- poj 3292 Semi-prime H-numbers
- POJ 3292 Semi-prime H-numbers
- Mathematics:Semi-prime H-numbers(POJ 3292)
- poj 3292 Semi-prime H-numbers (筛选法思想)
- POJ 3292 Semi-prime H-numbers (数论)
- POJ 3292 Semi-prime H-numbers