Optimization of sieve of Erathosthenes

#include 
    #include 
    #include 
    #include 
    using namespace std;
    #define RUNS 1000
    char z[100000];
    int i,j,k,c;

    void main(void)
      {
      DWORD starttime,endtime;
      float totaltime;

      starttime = GetTickCount();//get start time
      for(k=0;k<RUNS;k++)
      {
          c=0; 

      //for (i=0;i<10000;i++) z[i]=0; //clear array
      memset(z,0,100000);
        z[0]=1; // 0 is not a prime
        z[1]=1;  // 1 is not a prime
            //now remove all evens from 4 up
            for(i=4;i<100000;i=i+2) z[i]=1; //remove evens
            //now loop through remaing up to square root of max number
            for(i=3;i<316;i=i+2)
            {
            if(z[i]==0) for(j=2*i;j<100000;j=j+i) z[j]=1;
            }
      }

      endtime=GetTickCount();//get finish time
      //calc time
      for(i=0;i<100000;i++)
      {
      if(z[i]==0) {cout<<i<<" ";c++;}
      }
      cout<<"primes found="<<c<<endl;
      totaltime=((float)endtime-(float)starttime)/(1000.0*RUNS);//calculate total time in secs
      cout<<"Totaltime="<<totaltime<<" sec\n";

      printf("Press any key to end");

      getch();
      }

• I'm trying to find any optimization for my sieve of Eratosthenes code for counting first 100000 prime numbers.

• The program first mark all even numbers, than square root of max number.

• The program already does take fraction of the seconds to count these prime numbers, but I`m looking for any optimization to make it even quicker.

if (z[i] == 0)
    for (j = 2 * i; j < 100000; j = j + i)
        z[j] = 1;

if (z[i] == 0)
    for (j = i * i; j < 100000; j = j + 2 * i)
        z[j] = 1;

• k * i (with k

• when j is odd, j + i` is even and so already set by even sieve.

const int N = 100000;
const int SQRT_N = 316;

void shieve(char (&isNotPrime)[N])
{
    memset(isNotPrime, 0, N);
    isNotPrime[0] = 1; // 0 is not a prime
    isNotPrime[1] = 1; // 1 is not a prime
    // now remove all evens from 4 up
    for (int i = 4; i < N; i = i + 2) {
        isNotPrime[i] = 1; // remove evens
    }
    // now loop through remaing up to square root of max number
    for (int i = 3; i < SQRT_N; i = i + 2) {
        if (!isNotPrime[i]) {
            for(int j = i * i; j < N; j = j + 2 * i) {
                isNotPrime[j] = 1;
            }
        }
    }
}