用c++找出质数最快的算法是什么?我已经使用了sieve的算法,但我仍然希望它更快!
当前回答
#include <iostream>
using namespace std;
int set [1000000];
int main (){
for (int i=0; i<1000000; i++){
set [i] = 0;
}
int set_size= 1000;
set [set_size];
set [0] = 2;
set [1] = 3;
int Ps = 0;
int last = 2;
cout << 2 << " " << 3 << " ";
for (int n=1; n<10000; n++){
int t = 0;
Ps = (n%2)+1+(3*n);
for (int i=0; i==i; i++){
if (set [i] == 0) break;
if (Ps%set[i]==0){
t=1;
break;
}
}
if (t==0){
cout << Ps << " ";
set [last] = Ps;
last++;
}
}
//cout << last << endl;
cout << endl;
system ("pause");
return 0;
}
其他回答
#include<stdio.h>
main()
{
long long unsigned x,y,b,z,e,r,c;
scanf("%llu",&x);
if(x<2)return 0;
scanf("%llu",&y);
if(y<x)return 0;
if(x==2)printf("|2");
if(x%2==0)x+=1;
if(y%2==0)y-=1;
for(b=x;b<=y;b+=2)
{
z=b;e=0;
for(c=2;c*c<=z;c++)
{
if(z%c==0)e++;
if(e>0)z=3;
}
if(e==0)
{
printf("|%llu",z);
r+=1;
}
}
printf("|\n%llu outputs...\n",r);
scanf("%llu",&r);
}
我会让你决定这是不是最快的。
using System;
namespace PrimeNumbers
{
public static class Program
{
static int primesCount = 0;
public static void Main()
{
DateTime startingTime = DateTime.Now;
RangePrime(1,1000000);
DateTime endingTime = DateTime.Now;
TimeSpan span = endingTime - startingTime;
Console.WriteLine("span = {0}", span.TotalSeconds);
}
public static void RangePrime(int start, int end)
{
for (int i = start; i != end+1; i++)
{
bool isPrime = IsPrime(i);
if(isPrime)
{
primesCount++;
Console.WriteLine("number = {0}", i);
}
}
Console.WriteLine("primes count = {0}",primesCount);
}
public static bool IsPrime(int ToCheck)
{
if (ToCheck == 2) return true;
if (ToCheck < 2) return false;
if (IsOdd(ToCheck))
{
for (int i = 3; i <= (ToCheck / 3); i += 2)
{
if (ToCheck % i == 0) return false;
}
return true;
}
else return false; // even numbers(excluding 2) are composite
}
public static bool IsOdd(int ToCheck)
{
return ((ToCheck % 2 != 0) ? true : false);
}
}
}
在我使用2.40 GHz处理器的酷睿2 Duo笔记本电脑上,查找并打印1到1,000,000范围内的质数大约需要82秒。它找到了78,498个质数。
这是我一直在玩的埃拉托色尼筛子的Python实现。
def eratosthenes(maximum: int) -> list[int | None]:
"""
Find all the prime numbers between 2 and `maximum`.
Args:
maximum: The maximum number to check.
Returns:
A list of primes between 2 and `maximum`.
"""
if maximum < 2:
return []
# Discard even numbers by default.
sequence = dict.fromkeys(range(3, maximum+1, 2), True)
for num, is_prime in sequence.items():
# Already filtered, let's skip it.
if not is_prime:
continue
# Avoid marking the same number twice.
for num2 in range(num ** 2, maximum+1, num):
# Here, `num2` might contain an even number - skip it.
if num2 in sequence:
sequence[num2] = False
# Re-add 2 as prime and filter out the composite numbers.
return [2] + [num for num, is_prime in sequence.items() if is_prime]
在一台简陋的三星Galaxy A40上,该代码大约需要16秒才能输入10000000个数字。
欢迎提出建议!
#include <iostream>
using namespace std;
int set [1000000];
int main (){
for (int i=0; i<1000000; i++){
set [i] = 0;
}
int set_size= 1000;
set [set_size];
set [0] = 2;
set [1] = 3;
int Ps = 0;
int last = 2;
cout << 2 << " " << 3 << " ";
for (int n=1; n<10000; n++){
int t = 0;
Ps = (n%2)+1+(3*n);
for (int i=0; i==i; i++){
if (set [i] == 0) break;
if (Ps%set[i]==0){
t=1;
break;
}
}
if (t==0){
cout << Ps << " ";
set [last] = Ps;
last++;
}
}
//cout << last << endl;
cout << endl;
system ("pause");
return 0;
}
这是找到从1到n的所有质数的最快算法(在我的电脑上,它只花了0.004秒就找到了从1到1000000的所有质数)。
#include <iostream>
#include <fstream>
using namespace std;
double FindPrime(bool* array, int size){
clock_t start;
double runtime;
for (int i = 2; i < size; i++)
array[i] = true;
start = clock();
for (int i = 2; i <= size; i++)
if (array[i])
for (int j = 2 * i; j < size; j += i)
array[j] = false;
runtime = (double)(clock() - start) / CLOCKS_PER_SEC;
return runtime;
}
int main() {
ofstream fout("prime.txt");
int n = 0;
cout << "Enter the upper limit of prime numbers searching algorithm:";
cin >> n;
bool* array = new bool[n + 1];
double duration = FindPrime(array, n + 1);
printf("\n%f seconds.\n", duration);
for (int i = 2; i <= n; i++)
if (array[i])
fout << i << endl;
fout.close();
return 0;
}