A more common judgment method
bool IsPrime()
{
for(int i=2;i<=sqrt(n);i++)
{
if(n%2==0)
return false;
}
return true;
}
Eratosthenes screening method (multiples of prime numbers must not be prime numbers)
At the same time, for each x, the multiple of x greater than or equal to the square of x is marked as a composite number.
void primes(int n)
{
memset(v,0,sizeof(v));
for(i=2;i<=n;i++)
{
if(v[i]) continue;
cout<<i<<endl;
for(j=i;j<=n/i;j++)
v[i][j]=1;
}
}
Linear sieve method
Each composite number will only be sifted by its smallest prime factor once, and the time complexity is O(n)
int v[maxn],prime[maxn];
void primes(int n)
{
memset(v,0,sizeof(v));///最小质因子
m=0;
for(i=2;i<=n;i++)
{
if(v[i]==0)///i为质数
{
v[i]=i;
prime[++m]=i;
}
///给当前的数i乘上一个质因子
for(j=1;j<=m;j++)
{
///i有比prime[i]更小的质因子，或者超出n的范围
if(prime[j]>v[i]||prime[j]>n/i) break;
v[i*prime[j]]=prime[j];
}
}
for(i=1;i<=m;i++)
cout<<prime[i]<<endl;
}
There is also a method for judging prime numbers that I saw on the Internet. This was a template that was also used by ccpc at that time.
bool isPrime( int num )
{
//两个较小数另外处理
if(num ==2|| num==3 )
return 1 ;
//不在6的倍数两侧的一定不是质数
if(num %6!= 1&&num %6!= 5)
return 0 ;
int tmp =sqrt( num);
//在6的倍数两侧的也可能不是质数
for(int i= 5;i <=tmp; i+=6 )
if(num %i== 0||num %(i+ 2)==0 )
return 0 ;
//排除所有，剩余的是质数
return 1 ;
}
This is the fastest way I have seen to judge prime numbers