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Perfect numbers: Difference between revisions
→{{header|C}}: prime factors
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(→{{header|C}}: prime factors) |
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return 0;
}</lang>
Using functions from [[Factors of an integer#Prime factoring]]:<lang c>int main()
{
int j;
ulong fac[10000], n, sum;
sieve();
for (n = 2; n < 33550337; n++) {
j = get_factors(n, fac) - 1;
for (sum = 0; j && sum <= n; sum += fac[--j]);
if (sum == n) printf("%lu\n", n);
}
return 0;
}</lang>
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