stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.4661898074165522x):
#include <iostream>

#include <cstring>

#include <cmath>



using namespace std;





int main()

{

    unsigned int n;

    while( cin >> n )

    {

        bool *prime = new bool[n+2];

        unsigned int k, allprime = 0;



        memset( prime, false, n+2 );

        prime[2] = true;

        for( int i = 3; i <= n; i++ )

        {

            if( i % 2 == 0 )

            {

                prime[i] = false;

            }

            else

            {

                prime[i] = true;

            }

        }

        k = sqrt( n );

        for( int i = 3; i <= k; i++ )

        {

            for( int j = i + i; j <= n; j += i )

            {

                prime[j] = false;

            }

        }

        for( int i = 2; i <= n; i++ )

        {

            if( prime[i] )

            {

                allprime++;

            }

        }

        cout << allprime << '\n';

    }

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MAX 1000000

#define LMT 1000

unsigned flag[MAX>>6];

int primes[6000000];//5761455

#define ifc(n) (flag[n>>6]&(1<<((n>>1)&31)))

#define isc(n) (flag[n>>6]|=(1<<((n>>1)&31)))



int sieve()

{

    unsigned i, j, k;

    for(i=3; i<LMT; i+=2)

        if(!ifc(i))

            for(j=i*i, k=i<<1; j<MAX; j+=k)

                isc(j);

    return 0;

}



int main()

{

    sieve();

    unsigned int i,L;

    int k=1, count;

    primes[0]=2;

    while(scanf("%d", &L)==1)

    {

        count=1;

        for( i=3; i<=L; i+=2)

        {

            if(ifc(i)==0)

                count++;

        }

        if(L>1)

            printf("%d\n", count);

        else printf("0\n");

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.36938221491109x):
#include <iostream>

#include <cstring>

#include <cmath>



using namespace std;





int main()

{

    unsigned int n;

    while( cin >> n )

    {

        bool *prime = new bool[n+2];

        unsigned int k, allprime = 0;



        memset( prime, false, n+2 );

        prime[2] = true;

        for( int i = 3; i <= n; i++ )

        {

            if( i % 2 == 0 )

            {

                prime[i] = false;

            }

            else

            {

                prime[i] = true;

            }

        }

        k = sqrt( n );

        for( int i = 3; i <= k; i++ )

        {

            for( int j = i + i; j <= n; j += i )

            {

                prime[j] = false;

            }

        }

        for( int i = 2; i <= n; i++ )

        {

            if( prime[i] )

            {

                allprime++;

            }

        }

        cout << allprime << '\n';

    }

    return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.4664410483468915x):
#include<iostream>

#include<cmath>

using namespace std;

bool prime[1000000] = {false};

int main(){

    int n;

    while(cin >> n){

       

        prime[1] = true;

        prime[2] = false;

        int count = 0;

        

      

        for(int j = 2; j * j <= n ; j++){

            if(!prime[j]){

                for (int k = 2; k*j <= n;k++){

                   prime[j*k] = true;

                }

            }

        }

        for(int i = 1;i<=n;i++){

            if(!prime[i]) count = count +1;

        }

        cout << count << endl;

    }

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MAX 1000000

#define LMT 1000

unsigned flag[MAX>>6];

int primes[6000000];//5761455

#define ifc(n) (flag[n>>6]&(1<<((n>>1)&31)))

#define isc(n) (flag[n>>6]|=(1<<((n>>1)&31)))



int sieve()

{

    unsigned i, j, k;

    for(i=3; i<LMT; i+=2)

        if(!ifc(i))

            for(j=i*i, k=i<<1; j<MAX; j+=k)

                isc(j);

    return 0;

}



int main()

{

    sieve();

    unsigned int i,L;

    int k=1, count;

    primes[0]=2;

    while(scanf("%d", &L)==1)

    {

        count=1;

        for( i=3; i<=L; i+=2)

        {

            if(ifc(i)==0)

                count++;

        }

        if(L>1)

            printf("%d\n", count);

        else printf("0\n");

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.36955395559036x):
#include<iostream>

#include<cmath>

using namespace std;

bool prime[1000000] = {false};

int main(){

    int n;

    while(cin >> n){

       

        prime[1] = true;

        prime[2] = false;

        int count = 0;

        

      

        for(int j = 2; j * j <= n ; j++){

            if(!prime[j]){

                for (int k = 2; k*j <= n;k++){

                   prime[j*k] = true;

                }

            }

        }

        for(int i = 1;i<=n;i++){

            if(!prime[i]) count = count +1;

        }

        cout << count << endl;

    }

    return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.5353885293787364x):
#include<iostream>

#include<cstring>

#include<cmath>

using namespace std;



bool prime[1000000];



int main()

{

    int i,j,k,l,m,n;

    int count;

    while(cin>>n)

    {

        memset(prime,0,sizeof(prime));

        prime[1]=1;

        prime[2]=0;



        count=0;



        for(j=2;j*j<=n;j++)

        {

            if(prime[j]==0)

            {

                for(k=2;k*j<=n;k++)

                {

                   prime[j*k]=1;

                }

            }

        }

        for(i=1;i<=n;i++)

        {

            if(prime[i]==0)

            count++;

        }

        cout << count << endl;

    }

    return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.5254010611986182x):
#include<iostream>

#include<cstring>

#include<math.h>

using namespace std;

const int MAX = 1000000;

int prime(int n)

{

	int ans = 0;

	bool prime[MAX];

	memset(prime, 0, sizeof(prime));

	for (int i = 2; i < n;i++)  

	{ 

		for (int j = 2; j <= n / i; j++)

		{

			if (i*j <= n)

				prime[i*j] = 1;

		}

	}

	for (int i = 2; i <= n; i++)

	{

		if (!prime[i])

			ans++;

	}

	return ans;

}

int main()

{

	int n, m = 0;

	while (cin >> n&&m<30)

	{

		cout << prime(n) << endl;

		m++;

	}

	return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.5085245464259422x):


#include<iostream>

#include<cstdio>

#include<cstdlib>

using namespace std;



const int MAX_N = 999999;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n)

{

	int p = 0;

	for (int i = 0; i <= n; i++)

		is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;



	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i])

		{

			prime[p++] = i;

			for (int j = i * 2; j <= n; j += i)

				is_prime[j] = false;

		}

	}

		return p;

}



int main()

{

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MAX 1000000

#define LMT 1000

unsigned flag[MAX>>6];

int primes[6000000];//5761455

#define ifc(n) (flag[n>>6]&(1<<((n>>1)&31)))

#define isc(n) (flag[n>>6]|=(1<<((n>>1)&31)))



int sieve()

{

    unsigned i, j, k;

    for(i=3; i<LMT; i+=2)

        if(!ifc(i))

            for(j=i*i, k=i<<1; j<MAX; j+=k)

                isc(j);

    return 0;

}



int main()

{

    sieve();

    unsigned int i,L;

    int k=1, count;

    primes[0]=2;

    while(scanf("%d", &L)==1)

    {

        count=1;

        for( i=3; i<=L; i+=2)

        {

            if(ifc(i)==0)

                count++;

        }

        if(L>1)

            printf("%d\n", count);

        else printf("0\n");

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.398320958388763x):


#include<iostream>

#include<cstdio>

#include<cstdlib>

using namespace std;



const int MAX_N = 999999;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n)

{

	int p = 0;

	for (int i = 0; i <= n; i++)

		is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;



	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i])

		{

			prime[p++] = i;

			for (int j = i * 2; j <= n; j += i)

				is_prime[j] = false;

		}

	}

		return p;

}



int main()

{

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<cmath>

using namespace std;



bool prime[1000000];



int main()

{

    int i,j,k,l,m,n;

    int count;

    while(cin>>n)

    {

        memset(prime,0,sizeof(prime));

        prime[1]=1;

        prime[2]=0;



        count=0;



        for(j=2;j*j<=n;j++)

        {

            if(prime[j]==0)

            {

                for(k=2;k*j<=n;k++)

                {

                   prime[j*k]=1;

                }

            }

        }

        for(i=1;i<=n;i++)

        {

            if(prime[i]==0)

            count++;

        }

        cout << count << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.285105352353775x):


#include<iostream>

#include<cstdio>

#include<cstdlib>

using namespace std;



const int MAX_N = 999999;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n)

{

	int p = 0;

	for (int i = 0; i <= n; i++)

		is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;



	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i])

		{

			prime[p++] = i;

			for (int j = i * 2; j <= n; j += i)

				is_prime[j] = false;

		}

	}

		return p;

}



int main()

{

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<math.h>

using namespace std;

const int MAX = 1000000;

int prime(int n)

{

	int ans = 0;

	bool prime[MAX];

	memset(prime, 0, sizeof(prime));

	for (int i = 2; i < n;i++)  

	{ 

		for (int j = 2; j <= n / i; j++)

		{

			if (i*j <= n)

				prime[i*j] = 1;

		}

	}

	for (int i = 2; i <= n; i++)

	{

		if (!prime[i])

			ans++;

	}

	return ans;

}

int main()

{

	int n, m = 0;

	while (cin >> n&&m<30)

	{

		cout << prime(n) << endl;

		m++;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.3000669369333204x):


#include<iostream>

#include<cstdio>

#include<cstdlib>

using namespace std;



const int MAX_N = 999999;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n)

{

	int p = 0;

	for (int i = 0; i <= n; i++)

		is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;



	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i])

		{

			prime[p++] = i;

			for (int j = i * 2; j <= n; j += i)

				is_prime[j] = false;

		}

	}

		return p;

}



int main()

{

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

	return 0;

}
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.508465580874555x):
#include<iostream>

#include<cstdlib>

#include<algorithm>

#include<cstdio>

using namespace std;



//C



const int MAX_N = 1000000;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n){

	int p = 0;

	for (int i = 0; i <= n; i++)is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;

	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i]){

			prime[p++] = i;

			for (int j = 2 * i; j <= n; j += i)is_prime[j] = false;

		}

	}

	return p;

}



int main(){

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MAX 1000000

#define LMT 1000

unsigned flag[MAX>>6];

int primes[6000000];//5761455

#define ifc(n) (flag[n>>6]&(1<<((n>>1)&31)))

#define isc(n) (flag[n>>6]|=(1<<((n>>1)&31)))



int sieve()

{

    unsigned i, j, k;

    for(i=3; i<LMT; i+=2)

        if(!ifc(i))

            for(j=i*i, k=i<<1; j<MAX; j+=k)

                isc(j);

    return 0;

}



int main()

{

    sieve();

    unsigned int i,L;

    int k=1, count;

    primes[0]=2;

    while(scanf("%d", &L)==1)

    {

        count=1;

        for( i=3; i<=L; i+=2)

        {

            if(ifc(i)==0)

                count++;

        }

        if(L>1)

            printf("%d\n", count);

        else printf("0\n");

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.3982806513263943x):
#include<iostream>

#include<cstdlib>

#include<algorithm>

#include<cstdio>

using namespace std;



//C



const int MAX_N = 1000000;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n){

	int p = 0;

	for (int i = 0; i <= n; i++)is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;

	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i]){

			prime[p++] = i;

			for (int j = 2 * i; j <= n; j += i)is_prime[j] = false;

		}

	}

	return p;

}



int main(){

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<cmath>

using namespace std;



bool prime[1000000];



int main()

{

    int i,j,k,l,m,n;

    int count;

    while(cin>>n)

    {

        memset(prime,0,sizeof(prime));

        prime[1]=1;

        prime[2]=0;



        count=0;



        for(j=2;j*j<=n;j++)

        {

            if(prime[j]==0)

            {

                for(k=2;k*j<=n;k++)

                {

                   prime[j*k]=1;

                }

            }

        }

        for(i=1;i<=n;i++)

        {

            if(prime[i]==0)

            count++;

        }

        cout << count << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.285066948034439x):
#include<iostream>

#include<cstdlib>

#include<algorithm>

#include<cstdio>

using namespace std;



//C



const int MAX_N = 1000000;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n){

	int p = 0;

	for (int i = 0; i <= n; i++)is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;

	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i]){

			prime[p++] = i;

			for (int j = 2 * i; j <= n; j += i)is_prime[j] = false;

		}

	}

	return p;

}



int main(){

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<math.h>

using namespace std;

const int MAX = 1000000;

int prime(int n)

{

	int ans = 0;

	bool prime[MAX];

	memset(prime, 0, sizeof(prime));

	for (int i = 2; i < n;i++)  

	{ 

		for (int j = 2; j <= n / i; j++)

		{

			if (i*j <= n)

				prime[i*j] = 1;

		}

	}

	for (int i = 2; i <= n; i++)

	{

		if (!prime[i])

			ans++;

	}

	return ans;

}

int main()

{

	int n, m = 0;

	while (cin >> n&&m<30)

	{

		cout << prime(n) << endl;

		m++;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.300028281164102x):
#include<iostream>

#include<cstdlib>

#include<algorithm>

#include<cstdio>

using namespace std;



//C



const int MAX_N = 1000000;

int prime[MAX_N];

bool is_prime[MAX_N + 1];



int sieve(int n){

	int p = 0;

	for (int i = 0; i <= n; i++)is_prime[i] = true;

	is_prime[0] = is_prime[1] = false;

	for (int i = 2; i <= n; i++)

	{

		if (is_prime[i]){

			prime[p++] = i;

			for (int j = 2 * i; j <= n; j += i)is_prime[j] = false;

		}

	}

	return p;

}



int main(){

	int n;

	while (scanf("%d", &n) != EOF)

	{

		int ans = sieve(n);

		printf("%d\n", ans);

	}

}
// Slower C++ solution:
#include<stdio.h>

#include<math.h>

bool flag[1000000];

int main(void)

{

	long long i,j,p,n,count;

	p = (long long)sqrt(1000000.0);

	for(i = 3; i <= p; i = i + 2)

	{

		if(flag[i] == false)

		{

			for(j = i * i; j <= 1000000; j = j + 2 * i)

			{

				flag[j] = true;

				

			}

		}

	}

		

	while((scanf("%lld",&n)) != EOF)

	{

		

		if(n == 1 || n == 0)

			count = 0;

		else

		{

			count = 1;

			for(i = 3; i <= n; i = i + 2)

			{

				if(flag[i] == false)

					count++;

			}

		}

		printf("%lld\n",count);

	}

	



		

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.4780554701291133x):
#include<iostream>

#include<cstdio>

#include<cmath>

#define size 20000000

using namespace std;

int find_prime(int n)

{

    int count =0;

    bool *numberList=new bool[size+1];

    for(unsigned long i=3;i<=sqrt(n);i+=2)

    {

        for(unsigned long j=i;i*j<=n;j+=2)

            numberList[i*j]=true;

    }

    for (unsigned long i=3;i<=n;i+=2)

    {

        if(!numberList[i])

            count++;

    }



    return count;



}

int main ()

{

    int n;

    while((scanf("%d",&n))!=EOF)

    {

        if(n==1||n==0)

            printf("0\n");

        else if(n==2)

            printf("1\n");



        else

        {

        int v=0;

        v=find_prime(n);

        printf("%d\n",v+1);

        }

    }

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

#define MAX 1000000

#define LMT 1000

unsigned flag[MAX>>6];

int primes[6000000];//5761455

#define ifc(n) (flag[n>>6]&(1<<((n>>1)&31)))

#define isc(n) (flag[n>>6]|=(1<<((n>>1)&31)))



int sieve()

{

    unsigned i, j, k;

    for(i=3; i<LMT; i+=2)

        if(!ifc(i))

            for(j=i*i, k=i<<1; j<MAX; j+=k)

                isc(j);

    return 0;

}



int main()

{

    sieve();

    unsigned int i,L;

    int k=1, count;

    primes[0]=2;

    while(scanf("%d", &L)==1)

    {

        count=1;

        for( i=3; i<=L; i+=2)

        {

            if(ifc(i)==0)

                count++;

        }

        if(L>1)

            printf("%d\n", count);

        else printf("0\n");

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.3774932220287965x):
#include<iostream>

#include<cstdio>

#include<cmath>

#define size 20000000

using namespace std;

int find_prime(int n)

{

    int count =0;

    bool *numberList=new bool[size+1];

    for(unsigned long i=3;i<=sqrt(n);i+=2)

    {

        for(unsigned long j=i;i*j<=n;j+=2)

            numberList[i*j]=true;

    }

    for (unsigned long i=3;i<=n;i+=2)

    {

        if(!numberList[i])

            count++;

    }



    return count;



}

int main ()

{

    int n;

    while((scanf("%d",&n))!=EOF)

    {

        if(n==1||n==0)

            printf("0\n");

        else if(n==2)

            printf("1\n");



        else

        {

        int v=0;

        v=find_prime(n);

        printf("%d\n",v+1);

        }

    }

    return 0;

}