Trabb Pardo–Knuth algorithm: Difference between revisions

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@@ Line 3,020: / Line 3,020: @@
 ===Procedural===
+{{Works with|Python|3.10}}
-<lang python>def f(x):
+<lang python>import math
-    return abs(x) ** 0.5 + 5 * x**3
-def ask():
+def f(x):
-    return [float(y)
+    return math.sqrt(abs(x)) + 5 * x**3
-            for y in input('\n11 numbers: ').strip().split()[:11]]
+def ask_numbers(n=11):
+    print(f'Enter {n} numbers:')
+    return (float(input('>')) for _ in range(n))
 if __name__ == '__main__':
-    s = ask()
+    for x in ask_numbers().reverse():
+        if (result := f(x)) > 400:
-    s.reverse()
+            print(f'f({x}): overflow')
-    for x in s:
-        result = f(x)
-        if result > 400:
-            print(' %s:%s' % (x, "TOO LARGE!"), end='')
         else:
-            print(' %s:%s' % (x, result), end='')
+            print(f'f({x}) = {result}')</lang>
+{{out}}
-    print('')</lang>
+<pre>Enter 11 numbers:
-{{out|Sample output}}
+>1
-<pre>
+>532
-numbers: 1 2 3 4 5 6 7 8 9 10 11
+>465
-.0:TOO LARGE! 10.0:TOO LARGE! 9.0:TOO LARGE! 8.0:TOO LARGE! 7.0:TOO LARGE! 6.0:TOO LARGE! 5.0:TOO LARGE! 4.0:322.0 3.0:136.73205080756887 2.0:41.41421356237309 1.0:6.0</pre>
+>0
+>-8456
+>1
+>2
+>3
+>4
+>5
+>98465465
+f(98465465.0): overflow
+f(5.0): overflow
+f(4.0) = 322.0
+f(3.0) = 136.73205080756887
+f(2.0) = 41.41421356237309
+f(1.0) = 6.0
+f(-8456.0) = -3023186413988.0435
+f(0.0) = 0.0
+f(465.0): overflow
+f(532.0): overflow
+f(1.0) = 6.0</pre>
 =={{header|R}}==