Count in factors: Difference between revisions

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@@ Line 1,735: / Line 1,735: @@
 =={{header|Julia}}==
-<lang Julia>
+{{works with|Julia|0.6}}
-function factor_print{T<:Integer}(n::T)
-    const SEP = " \u00d7 "
-    -2 < n || return "-1"*SEP*factor_print(-n)
-    if isprime(n) || n < 2
-        return string(n)
-    end
-    a = T[]
-    for (k, v) in factor(n)
-        append!(a, k*ones(T, v))
-    end
-    sort!(a)
-    join(a, SEP)
-end
+<lang julia>using Primes
-lo = -4
+function strfactor(n::Integer)
-hi = 40
+    n > -2 || return "-1 × " * strfactor(-n)
-println("Factor print ", lo, " to ", hi)
+    isprime(n) || n < 2 && return dec(n)
-for i in lo:hi
+    f = factor(Vector{typeof(n)}, n)
-    println(@sprintf("%5d = ", i), factor_print(i))
+    return join(f, " × ")
 end
-</lang>
+lo, hi = -4, 40
-I wrote this solution's <tt>factor_print</tt> function with ease rather than efficiency in mind.  It may be more efficient to first sort the keys of the <tt>factor</tt> dictionary and to build the string in place, but I find the logic of the presented solution to be clearer.  The <tt>factor</tt> built-in is relevant to this solution only for integers greater than <tt>1</tt>, but I've constructed <tt>factor_print</tt> to return meaningful results for any representable proper integer.
+println("Factor print $lo to $hi:")
+for n in lo:hi
+    @printf("%5d = %s\n", n, strfactor(n))
+end</lang>
 {{out}}
+<pre>Factor print -4 to 40:
-<pre>
-Factor print -4 to 40
    -4 = -1 × 2 × 2
    -3 = -1 × 3
@@ Line 1,806: / Line 1,797: @@
 = 2 × 19
 = 3 × 13
-= 2 × 2 × 2 × 5
+= 2 × 2 × 2 × 5</pre>
-</pre>
 =={{header|Kotlin}}==