Largest palindrome product: Difference between revisions
m
→{{header|Ring}}: keeping it odd
m (→{{header|Ring}}: removed library dependence, improved performance by checking hi to low and terminating the inner loop once it's obvious the result won't be improved) |
m (→{{header|Ring}}: keeping it odd) |
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Line 156:
// the best-found-so-far second number. Doing this
// lowers the iteration count by a lot.
for m = max to second step -
prod = n * m
if isPal(prod)
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