C++ 寻找区间素数的初始方法
在尝试查找范围内的素数时,我遇到了以下代码: 代码取自 然而,我觉得这段代码并不直观,也不容易理解 有人能解释一下上述算法背后的基本思想吗? 有哪些替代算法可以标记某个范围内的非素数?C++ 寻找区间素数的初始方法,c++,primes,sieve-of-eratosthenes,C++,Primes,Sieve Of Eratosthenes,在尝试查找范围内的素数时,我遇到了以下代码: 代码取自 然而,我觉得这段代码并不直观,也不容易理解 有人能解释一下上述算法背后的基本思想吗? 有哪些替代算法可以标记某个范围内的非素数? 这太复杂了。让我们从一个基本的Eratosthenes筛开始,在伪代码中,它输出所有小于或等于n的素数: 这个函数调用每个素数p的输出;输出可以打印素数,或对素数求和,或对它们进行计数,或对它们执行任何操作。外for循环依次考虑每个候选素数;筛分发生在内部for循环中,在该循环中,当前素数p的倍数从筛子中移除 一
这太复杂了。让我们从一个基本的Eratosthenes筛开始,在伪代码中,它输出所有小于或等于n的素数: 这个函数调用每个素数p的输出;输出可以打印素数,或对素数求和,或对它们进行计数,或对它们执行任何操作。外for循环依次考虑每个候选素数;筛分发生在内部for循环中,在该循环中,当前素数p的倍数从筛子中移除
一旦你了解了它是如何工作的,就开始讨论一个范围内的埃拉托斯烯的分段筛。你有没有从位的角度考虑过筛,它可以提供更多的素数,通过使用缓冲区,你可以修改它,例如使用64位整数,找到2到2^60之间的素数,同时保留已经发现的素数的偏移量。下面将使用整数数组 减退
基本原理简单;一旦你找到一个素数,这个数的所有倍数都不是素数。
// For each prime in sqrt(N) we need to use it in the segmented sieve process.
for (i = 0; i < cnt; i++) {
p = myPrimes[i]; // Store the prime.
s = M / p;
s = s * p; // The closest number less than M that is composite number for this prime p.
for (int j = s; j <= N; j = j + p) {
if (j < M) continue; // Because composite numbers less than M are of no concern.
/* j - M = index in the array primesNow, this is as max index allowed in the array
is not N, it is DIFF_SIZE so we are storing the numbers offset from.
while printing we will add M and print to get the actual number. */
primesNow[j - M] = false;
}
}
// In this loop the first prime numbers for example say 2, 3 are also set to false.
for (int i = 0; i < cnt; i++) { // Hence we need to print them in case they're in range.
if (myPrimes[i] >= M && myPrimes[i] <= N) // Without this loop you will see that for a
// range (1, 30), 2 & 3 doesn't get printed.
cout << myPrimes[i] << endl;
}
// primesNow[] = false for all composite numbers, primes found by checking with true.
for (int i = 0; i < N - M + 1; ++i) {
// i + M != 1 to ensure that for i = 0 and M = 1, 1 is not considered a prime number.
if (primesNow[i] == true && (i + M) != 1)
cout << i + M << endl; // Print our prime numbers in the range.
}
function primes(n)
sieve := makeArray(2..n, True)
for p from 2 to n
if sieve[p]
output(p)
for i from p*p to n step p
sieve[p] := False
#include <math.h> // sqrt(), the upper limit need to eliminate
#include <stdio.h> // for printing, could use <iostream>
#define BIT_SET(d, n) (d[n>>5]|=1<<(n-((n>>5)<<5)))
#define BIT_GET(d, n) (d[n>>5]&1<<(n-((n>>5)<<5)))
#define BIT_FLIP(d, n) (d[n>>5]&=~(1<<(n-((n>>5)<<5))))
unsigned int n = 0x80000; // the upper limit 1/2 mb, with 32 bits each
// will get the 1st primes upto 16 mb
int *data = new int[n]; // allocate
unsigned int r = n * 0x20; // the actual number of bits avalible
for(int i=0;i<n;i++)
data[i] = 0xFFFFFFFF;
unsigned int seed = 2; // the seed starts at 2
unsigned int uLimit = sqrt(r); // the upper limit for checking off the sieve
BIT_FLIP(data, 1); // one is not prime
// untill uLimit is reached
while(seed < uLimit) {
// don't include itself when eliminating canidates
for(int i=seed+seed;i<r;i+=seed)
BIT_FLIP(data, i);
// find the next bit still active (set to 1), don't include the current seed
for(int i=seed+1;i<r;i++) {
if (BIT_GET(data, i)) {
seed = i;
break;
}
}
}
unsigned long bit_index = 0; // the current bit
int w = 8; // the width of a column
unsigned pc = 0; // prime, count, to assist in creating columns
for(int i=0;i<n;i++) {
unsigned long long int b = 1; // double width, so there is no overflow
// if a bit is still set, include that as a result
while(b < 0xFFFFFFFF) {
if (data[i]&b) {
printf("%8.u ", bit_index);
if(((pc++) % w) == 0)
putchar('\n'); // add a new row
}
bit_index++;
b<<=1; // multiply by 2, to check the next bit
}
}
delete [] data;