C++编程求出3～100内的素数，求大神！！！！！！

Answer 1

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int N; cin>>N;
int *Location=new int[N+1];
for(int i=0;i!=N+1;++i)
Location[i]=i;
Location[1]=0;
int end=sqrt((double)N)+1;
for(int p=2;p!=end;++p)
{
if(Location[p])
{
for(int q=p;p*q<=N;++q)
{
if(Location[q])
{
for(int k=p*q;k<=N;k*=p)
Location[k]=0;
}
}
}
}
int m=0;
for(int i=1;i!=N+1;++i)
{
if(Location[i]!=0)
{
cout<<Location[i]<<" ";
++m;
}
if(m%10==0) cout<<endl;
}
cout<<endl<<m<<endl;
return 0;
}
重新改了一下，你再试试看，这是筛法求素数的过程，先输入上限，然后自动输出小于该上限的素数，并统计总数输出

Answer 2

问题解决，请采纳！
#include <iostream>
using namespace std;
bool isPrime(unsigned int n)
{
for(unsigned int i=2; i< n; i++)
{
if(n%i == 0)
return false;
}
return true;
}
void main()
{
for (int i =3; i<=100; ++i)
{
if (isPrime(i))
{
cout<<"this is a prime :"<<i<<endl;
}
}
};

Answer 3

#include <iostream>
#include <vector>

int main()
{
std::vector<bool> prime(101, true);
prime[0] = prime[1] = false;

for (int i = 2; i <= 10; i++) {
if (prime[i]) {
int t = i*i;
while(t<=100) {
prime[t] = false;
t += i;
}
}
}

for (int i = 1; i <= 100; i++) {
if (prime[i])
std::cout<<i<<" ";
}
std::cout<<std::endl;

return 0;
}

Answer 4

#include <iostream>
using namespace std;
bool IsPrime(int num)
{
    bool bIsPrime = true;

    if (num == 2)
    {
        return bIsPrime;
    }

    for (int i=2; i<=num-1; i++)
    {
        if ( num % i == 0 )
        {
            bIsPrime = false;
            break;
        }
    }

    return bIsPrime;
}

int _tmain(int argc, _TCHAR* argv[])
{
    for (int i=3; i<=100; i++)
    {
        if (IsPrime(i))
        {
            cout<<i<<" is a prime!"<<endl;
        }
    }

    return 0;
}