Legendre prime counting function: Difference between revisions
Added Java solution
(Added C++ solution) |
(Added Java solution) |
||
Line 284:
10^8 5761455
10^9 50847534
</pre>
=={{header|Java}}==
<lang java>import java.util.*;
public class LegendrePrimeCounter {
public static void main(String[] args) {
LegendrePrimeCounter counter = new LegendrePrimeCounter(1000000000);
for (int i = 0, n = 1; i < 10; ++i, n *= 10)
System.out.printf("10^%d\t%d\n", i, counter.primeCount((n)));
}
private List<Integer> primes;
private Map<Integer, Map<Integer, Integer>> phiCache = new HashMap<>();
public LegendrePrimeCounter(int limit) {
primes = generatePrimes((int)Math.sqrt((double)limit));
}
public int primeCount(int n) {
if (n < 2)
return 0;
int a = primeCount((int)Math.sqrt((double)n));
return phi(n, a) + a - 1;
}
private int phi(int x, int a) {
if (a == 0)
return x;
Map<Integer, Integer> map = phiCache.computeIfAbsent(x, k -> new HashMap<>());
Integer value = map.get(a);
if (value != null)
return value;
int result = phi(x, a - 1) - phi(x / primes.get(a - 1), a - 1);
map.put(a, result);
return result;
}
private static List<Integer> generatePrimes(int limit) {
boolean[] sieve = new boolean[limit >> 1];
Arrays.fill(sieve, true);
for (int p = 3, s = 9; s < limit; p += 2) {
if (sieve[p >> 1]) {
for (int q = s; q < limit; q += p << 1)
sieve[q >> 1] = false;
}
s += (p + 1) << 2;
}
List<Integer> primes = new ArrayList<>();
if (limit > 2)
primes.add(2);
for (int i = 1; i < sieve.length; ++i) {
if (sieve[i])
primes.add((i << 1) + 1);
}
return primes;
}
}</lang>
{{out}}
<pre>
10^0 0
10^1 4
10^2 25
10^3 168
10^4 1229
10^5 9592
10^6 78498
10^7 664579
10^8 5761455
10^9 50847534
</pre>
| |||