求len之内的所有的素数
除了比较常用的开根号的求法，还有两种更好的方法，埃氏筛和线性筛。其中埃氏筛更好理解，而线性筛（欧拉筛）不好理解但是更快。
埃氏筛
#include <bits/stdc++.h>
using namespace std;
int main()
{
int len = 100;
vector<int> primes;
bool isPrime[len+2];
memset(isPrime, 1, len);
for (int i = 2; i <= len; ++i)
{
if (isPrime[i])
{
primes.push_back(i);
for (int j = i*i; j <= len; j += i)
isPrime[j] = false;
}
}
for (vector<int>::iterator iter = primes.begin(); iter != primes.end(); ++iter)
cout<<*iter<<" ";
cout<<endl;
}
结果为：
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
讲解：
vector<int> primes用于放置质数，isPrime用于判断是否是质数，初始全为true，也就是全是质数。之后设置把2和3放进primes里面。- 然后
i从2开始，遍历到len，如果i是质数，就放进primes里，同时假如i是质数，那么所有i的倍数，肯定就不是质数了。所以2*i、3*i、4*i……都不是质数。所以我们就可以不断将这些数的isPrime设置为false。 - 但是我们的循环是从
i*i开始的，这是因为2*i、3*i到(i-1)*i的所有合数，肯定会在之前就被设置为false了。直观上看确实是，比如3*3是9,8会在4*2时被判断为合数，5*5是25,24会在8*3时被判断为合数。 - 具体证明不会，可以看这个。
欧拉筛（线性筛）
埃氏筛会有重复的情况，比如说40吧，在判断到2时，202是40，判断40是合数，在判断到5时，85=40，也判断40是合数，所以20就被判断了两次。这就有了欧拉筛。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int len = 100;
vector<int> primes;
bool isPrime[len+2];
memset(isPrime, 1, len);
for (int i = 2; i <= len; ++i)
{
if (isPrime[i] == true)
primes.push_back(i);
for (int j = 0; j < primes.size(); ++j)
{
if (i * primes[j] > len)
break;
isPrime[i * primes[j]] = false;
if (i % primes[j] == 0)
break;
}
}
for (vector<int>::iterator iter = primes.begin(); iter != primes.end(); ++iter)
cout<<*iter<<" ";
cout<<endl;
}
vector<int> primes用于放置质数，isPrime用于判断是否是质数，初始全为true，也就是全是质数。之后设置把2和3放进primes里面。- 然后
i从2开始，遍历到len，如果i是质数，就放进primes里。同时，我们对现在判断好的所有质数primes[j]，开始遍历，那么i*primes[j]肯定就不是质数了，设置其isPrime为false。 - 但是在遍历
primes[j]的时候，如果i % primes[j] == 0，那么就不用再看接下来的i*primes[j+1]、i*primes[j+2]……了，直接break。这是因为如果i / primes[j] = m，那么i = m * primes[j]，那么i*primes[j+1] = m * primes[j] * primes[j+1] = (m*primes[j+1])*primes[j] = n*primes[j]，也就是说这个数会在i*primes[j+1]的时候被判断，也会在n*primes[j]的时候被判断，其中n是肯定大于i的，因为i*primes[j+1] = n*primes[j]，但是primes[j+1] > primes[j]。那我们就等到当i遍历到n时，再去把i*primes[j+1]这个数给判断成合数。 - 我们可以把每次筛出来的合数打印出来：
i is 3, primes[j] is 2, multi is 6
i is 3, primes[j] is 3, multi is 9
i is 4, primes[j] is 2, multi is 8
i is 5, primes[j] is 2, multi is 10
i is 5, primes[j] is 3, multi is 15
i is 5, primes[j] is 5, multi is 25
i is 6, primes[j] is 2, multi is 12
i is 7, primes[j] is 2, multi is 14
i is 7, primes[j] is 3, multi is 21
i is 7, primes[j] is 5, multi is 35
i is 7, primes[j] is 7, multi is 49
i is 8, primes[j] is 2, multi is 16
i is 9, primes[j] is 2, multi is 18
i is 9, primes[j] is 3, multi is 27
i is 10, primes[j] is 2, multi is 20
i is 11, primes[j] is 2, multi is 22
i is 11, primes[j] is 3, multi is 33
i is 11, primes[j] is 5, multi is 55
i is 11, primes[j] is 7, multi is 77
i is 12, primes[j] is 2, multi is 24
i is 13, primes[j] is 2, multi is 26
i is 13, primes[j] is 3, multi is 39
i is 13, primes[j] is 5, multi is 65
i is 13, primes[j] is 7, multi is 91
i is 14, primes[j] is 2, multi is 28
i is 15, primes[j] is 2, multi is 30
i is 15, primes[j] is 3, multi is 45
i is 16, primes[j] is 2, multi is 32
i is 17, primes[j] is 2, multi is 34
i is 17, primes[j] is 3, multi is 51
i is 17, primes[j] is 5, multi is 85
i is 18, primes[j] is 2, multi is 36
i is 19, primes[j] is 2, multi is 38
i is 19, primes[j] is 3, multi is 57
i is 19, primes[j] is 5, multi is 95
i is 20, primes[j] is 2, multi is 40
i is 21, primes[j] is 2, multi is 42
i is 21, primes[j] is 3, multi is 63
i is 22, primes[j] is 2, multi is 44
i is 23, primes[j] is 2, multi is 46
i is 23, primes[j] is 3, multi is 69
i is 24, primes[j] is 2, multi is 48
i is 25, primes[j] is 2, multi is 50
i is 25, primes[j] is 3, multi is 75
i is 26, primes[j] is 2, multi is 52
i is 27, primes[j] is 2, multi is 54
i is 27, primes[j] is 3, multi is 81
i is 28, primes[j] is 2, multi is 56
i is 29, primes[j] is 2, multi is 58
i is 29, primes[j] is 3, multi is 87
i is 30, primes[j] is 2, multi is 60
i is 31, primes[j] is 2, multi is 62
i is 31, primes[j] is 3, multi is 93
i is 32, primes[j] is 2, multi is 64
i is 33, primes[j] is 2, multi is 66
i is 33, primes[j] is 3, multi is 99
i is 34, primes[j] is 2, multi is 68
i is 35, primes[j] is 2, multi is 70
i is 36, primes[j] is 2, multi is 72
i is 37, primes[j] is 2, multi is 74
i is 38, primes[j] is 2, multi is 76
i is 39, primes[j] is 2, multi is 78
i is 40, primes[j] is 2, multi is 80
i is 41, primes[j] is 2, multi is 82
i is 42, primes[j] is 2, multi is 84
i is 43, primes[j] is 2, multi is 86
i is 44, primes[j] is 2, multi is 88
i is 45, primes[j] is 2, multi is 90
i is 46, primes[j] is 2, multi is 92
i is 47, primes[j] is 2, multi is 94
i is 48, primes[j] is 2, multi is 96
i is 49, primes[j] is 2, multi is 98
i is 50, primes[j] is 2, multi is 100
每个合数都只被筛了一次
总结
- 两个方法都是先假设所有数是质数，然后从里面筛出来合数。
- 埃氏筛是每次得到一个质数时，就对该质数的所有倍数筛除，因为他们肯定是合数。
- 欧拉筛是每次
i从2到n判断其是否是质数时，就对现在确定的所有质数k遍历，把i*k筛掉。 - 但是欧拉筛会在必要的时候
break，保证每次筛掉的和数i*k，质数k都是最小的那个质数。比如20可以等于4*5，也可以是10*2，在这里i分别是4和10，质数k分别是5和2，我们会在10*2这里将20筛掉，而是在4*5这里，因为2是最小的那个质数。