桥头堡
哎哎哎:# t0]https://www . geeksforgeeks . org/pierppont-prime/
A 皮尔庞特素数是 p = 2l3k+1 形式的素数。前几个皮尔庞特素数是 2，3，5，7，13，17，19，37，73，97，109，… 给定一个数 n ，任务是打印小于 n 的皮尔庞特素数 示例:
Input : n = 15
Output : 2 3 5 7 13
Input : n = 200
Output : 2 3 5 7 13 17 19 37
73 97 109 163 193
这个想法是找到只有 2 和 3 的幂的数。现在用厄拉多塞筛求所有质数。最后，打印两个序列的公共编号。 以下是本办法的实施情况:
C++
// CPP program to print Pierpont prime
// numbers smaller than n.
#include <bits/stdc++.h>
using namespace std;
bool printPierpont(int n)
{
// Finding all numbers having factor power
// of 2 and 3 Using sieve
bool arr[n+1];
memset(arr, false, sizeof arr);
int two = 1, three = 1;
while (two + 1 < n) {
arr[two] = true;
while (two * three + 1 < n) {
arr[three] = true;
arr[two * three] = true;
three *= 3;
}
three = 1;
two *= 2;
}
// Storing number of the form 2^i.3^k + 1.
vector<int> v;
for (int i = 0; i < n; i++)
if (arr[i])
v.push_back(i + 1);
// Finding prime number using sieve of
// Eratosthenes. Reusing same array as
// result of above computations in v.
memset(arr, false, sizeof arr);
for (int p = 2; p * p < n; p++) {
if (arr[p] == false)
for (int i = p * 2; i < n; i += p)
arr[i] = true;
}
// Printing n pierpont primes smaller than n
for (int i = 0; i < v.size(); i++)
if (!arr[v[i]])
cout << v[i] << " ";
}
// Driven Program
int main()
{
int n = 200;
printPierpont(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print Pierpont prime
// numbers smaller than n.
import java.util.*;
class GFG{
static void printPierpont(int n)
{
// Finding all numbers having factor power
// of 2 and 3 Using sieve
boolean[] arr=new boolean[n+1];
int two = 1, three = 1;
while (two + 1 < n) {
arr[two] = true;
while (two * three + 1 < n) {
arr[three] = true;
arr[two * three] = true;
three *= 3;
}
three = 1;
two *= 2;
}
// Storing number of the form 2^i.3^k + 1.
ArrayList<Integer> v=new ArrayList<Integer>();
for (int i = 0; i < n; i++)
if (arr[i])
v.add(i + 1);
// Finding prime number using sieve of
// Eratosthenes. Reusing same array as
// result of above computations in v.
arr=new boolean[n+1];
for (int p = 2; p * p < n; p++) {
if (arr[p] == false)
for (int i = p * 2; i < n; i += p)
arr[i] = true;
}
// Printing n pierpont primes smaller than n
for (int i = 0; i < v.size(); i++)
if (!arr[v.get(i)])
System.out.print(v.get(i)+" ");
}
// Driven Program
public static void main(String[] args)
{
int n = 200;
printPierpont(n);
}
}
// this code is contributed by mits
Python 3
# Python3 program to print Pierpont
# prime numbers smaller than n.
def printPierpont(n):
# Finding all numbers having factor
# power of 2 and 3 Using sieve
arr = [False] * (n + 1);
two = 1;
three = 1;
while (two + 1 < n):
arr[two] = True;
while (two * three + 1 < n):
arr[three] = True;
arr[two * three] = True;
three *= 3;
three = 1;
two *= 2;
# Storing number of the form 2^i.3^k + 1.
v = [];
for i in range(n):
if (arr[i]):
v.append(i + 1);
# Finding prime number using
# sieve of Eratosthenes.
# Reusing same array as result
# of above computations in v.
arr1 = [False] * (len(arr));
p = 2;
while (p * p < n):
if (arr1[p] == False):
for i in range(p * 2, n, p):
arr1[i] = True;
p += 1;
# Printing n pierpont primes
# smaller than n
for i in range(len(v)):
if (not arr1[v[i]]):
print(v[i], end = " ");
# Driver Code
n = 200;
printPierpont(n);
# This code is contributed by mits
C
// C# program to print Pierpont prime
// numbers smaller than n.
using System;
using System.Collections;
class GFG{
static void printPierpont(int n)
{
// Finding all numbers having factor power
// of 2 and 3 Using sieve
bool[] arr=new bool[n+1];
int two = 1, three = 1;
while (two + 1 < n) {
arr[two] = true;
while (two * three + 1 < n) {
arr[three] = true;
arr[two * three] = true;
three *= 3;
}
three = 1;
two *= 2;
}
// Storing number of the form 2^i.3^k + 1.
ArrayList v=new ArrayList();
for (int i = 0; i < n; i++)
if (arr[i])
v.Add(i + 1);
// Finding prime number using sieve of
// Eratosthenes. Reusing same array as
// result of above computations in v.
arr=new bool[n+1];
for (int p = 2; p * p < n; p++) {
if (arr[p] == false)
for (int i = p * 2; i < n; i += p)
arr[i] = true;
}
// Printing n pierpont primes smaller than n
for (int i = 0; i < v.Count; i++)
if (!arr[(int)v[i]])
Console.Write(v[i]+" ");
}
// Driven Program
static void Main()
{
int n = 200;
printPierpont(n);
}
}
// this code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print
// Pierpont prime numbers
// smaller than n.
function printPierpont($n)
{
// Finding all numbers
// having factor power
// of 2 and 3 Using sieve
$arr = array_fill(0, $n + 1, false);
$two = 1;
$three = 1;
while ($two + 1 < $n)
{
$arr[$two] = true;
while ($two * $three + 1 < $n)
{
$arr[$three] = true;
$arr[$two * $three] = true;
$three *= 3;
}
$three = 1;
$two *= 2;
}
// Storing number of the
// form 2^i.3^k + 1.
$v;
$x = 0;
for ($i = 0; $i < $n; $i++)
if ($arr[$i])
$v[$x++] = $i + 1;
// Finding prime number using
// sieve of Eratosthenes.
// Reusing same array as result
// of above computations in v.
$arr1 = array_fill(0, count($arr), false);
for ($p = 2;
$p * $p < $n; $p++)
{
if ($arr1[$p] == false)
for ($i = $p * 2;
$i < $n; $i += $p)
$arr1[$i] = true;
}
// Printing n pierpont
// primes smaller than n
for ($i = 0; $i < $x; $i++)
if (!$arr1[$v[$i]])
echo $v[$i] . " ";
}
// Driver Code
$n = 200;
printPierpont($n);
// This Code is contributed by mits
?>
java 描述语言
<script>
// Javascript program to print Pierpont prime
// numbers smaller than n.
function printPierpont(n)
{
// Finding all numbers having factor power
// of 2 and 3 Using sieve
var arr = Array(n+1).fill(false);
var two = 1, three = 1;
while (two + 1 < n) {
arr[two] = true;
while (two * three + 1 < n) {
arr[three] = true;
arr[two * three] = true;
three *= 3;
}
three = 1;
two *= 2;
}
// Storing number of the form 2^i.3^k + 1.
var v = [];
for (var i = 0; i < n; i++)
if (arr[i])
v.push(i + 1);
// Finding prime number using sieve of
// Eratosthenes. Reusing same array as
// result of above computations in v.
arr = Array(n+1).fill(false);
for (var p = 2; p * p < n; p++) {
if (arr[p] == false)
for (var i = p * 2; i < n; i += p)
arr[i] = true;
}
// Printing n pierpont primes smaller than n
for (var i = 0; i < v.length; i++)
if (!arr[v[i]])
document.write( v[i] + " ");
}
// Driven Program
var n = 200;
printPierpont(n);
</script>
输出:
2 3 5 7 13 17 19 37 73 97 109 163 193
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