## Sunday, November 29, 2015

### C-code for the PrimePi function

This code uses the algorithm described here. The same algorithm has been used for SumOfPrimes function here.

#include<stdio.h>
#include<math.h>

#define M 1000000
#define Mod 1000000000

int Lower[M+1], Higher[M+1];
char Used[M+1];

long long SumNat(long long n)
{
long long p = n, q = n+1;
if(p%2)
q /= 2;
else
p /= 2;
return (p % Mod) * (q % Mod) % Mod;
}

long long PrimePi(long long n)
{
long long v = sqrt(n), p, temp, q, j, end, i, d, t;
Higher[1] = n % Mod - 1;
for(p = 2; p <= v; p++) {
Lower[p] = p - 1;
Higher[p] = (n/p) % Mod - 1;
}
for(p = 2; p <= v; p++) {
if(Lower[p] == Lower[p-1])
continue;
temp = Lower[p-1];
q = p * p;
Higher[1] -= Higher[p] - temp;
if(Higher[1] < 0)
Higher[1] += Mod;
j = 1 + (p & 1);
end = (v <= n/q) ? v : n/q;
for(i = p + j; i <= 1 + end; i += j) {
if(Used[i])
continue;
d = i * p;
if(d <= v)
Higher[i] -= Higher[d] - temp;
else {
t = n/d;
Higher[i] -= Lower[t] - temp;
}
if(Higher[i] < 0)
Higher[i] += Mod;
}
if(q <= v)
for(i = q; i <= end; i += p*j)
Used[i] = 1;
for(i = v; i >= q; i--) {
t = i/p;
Lower[i] -= Lower[t] - temp;
}
}
return Higher[1] % Mod;
}

int main() {
printf("%lld", PrimePi(1000000000000LL));
return 0;
}

Well, I conclude the discussion on the PrimePi functions with this post, I'll think of something to continue blogging which may take a while.

Until then
Yours Aye
Me