Multiplicatively perfect numbers: Difference between revisions

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=={{header|C}}==

This includes '1' as an MPN.

This includes '1' as an MPN. Run time around 3.2 seconds.

<syntaxhighlight lang="c">#include <stdio.h>

#include <stdbool.h>

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void divisors(int n, int *divs, int *length) {

if (n < 1) {

*length = 0;

return;

}

int i, j, k = 1, c = 0;

if (n%2) k = 2;

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if (!(n%i)) {

divs[c++] = i;

if (c > 3) break; // not eligible if has > 2 divisors

j = n / i;

if (j != i) divs[c++] = j;

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int main() {

int i, d, j, k, t, length, prod;

int divs[~~200~~], count = 0, limit = 500, s = 3, c = 3, squares = 1, cubes = 1;

int divs[5], count = 0, limit = 500, s = 3, c = 3, squares = 1, cubes = 1;

printf("Multiplicatively perfect numbers under %d:\n", limit);

setlocale(LC_NUMERIC, "");

for (i = 0; ; ++i) {

for (i = 1; ; ++i) {

if (i != 1) {

divisors(i, divs, &length);

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length = 2;

}

if (length > 1) {

if (length == 2 && divs[0] * divs[1] == i) {

~~prod = 1~~;

++count;

~~for~~ (~~d = 0; d~~ < ~~length; ++d~~) ~~prod *= divs[d];~~

if (i < 500) {

if (~~prod~~ == i) {

printf("%3d ", i);

++count;

if (!(count%10)) printf("\n");

if (i < 500) {

printf("%3d ", i);

if (!(count%10)) printf("\n");

}

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for (k = c; k * k * k < limit; k +=2 ) if (isPrime(k)) ++cubes;

t = count + squares - cubes - 1;

printf("Counts under %'7d: MPNs = %'7d Semi-primes = %'7d\n", limit, count, t);

printf("Counts under %'9d: MPNs = %'7d Semi-primes = %'7d\n", limit, count, t);

if (limit == ~~500000~~) break;

if (limit == 5000000) break;

s = j;

c = k;

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469 471 473 478 481 482 485 489 493 497

Counts under 500: MPNs = 150 Semi-primes = 153

Counts under 5,000: MPNs = 1,354 Semi-primes = 1,365

Counts under 50,000: MPNs = 12,074 Semi-primes = 12,110

Counts under 500,000: MPNs = 108,223 Semi-primes = 108,326

Counts under 5,000,000: MPNs = 978,983 Semi-primes = 979,274

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