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author | Timotej Lazar <timotej.lazar@fri.uni-lj.si> | 2016-01-13 13:51:15 +0100 |
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committer | Timotej Lazar <timotej.lazar@fri.uni-lj.si> | 2016-01-13 13:51:15 +0100 |
commit | 388b32039898cc8f1784378689007254eb7a33b6 (patch) | |
tree | ac9e4c6145ad323dc5479179a04f44bbcea73eff /prolog/problems/old_exams/pascal_3 | |
parent | e36a8a2feca7552d236c0f6e89ac73e7e690e7b1 (diff) |
Remove space-padding from <pre> blocks
Diffstat (limited to 'prolog/problems/old_exams/pascal_3')
-rw-r--r-- | prolog/problems/old_exams/pascal_3/en.py | 24 |
1 files changed, 12 insertions, 12 deletions
diff --git a/prolog/problems/old_exams/pascal_3/en.py b/prolog/problems/old_exams/pascal_3/en.py index bd17577..8481bd6 100644 --- a/prolog/problems/old_exams/pascal_3/en.py +++ b/prolog/problems/old_exams/pascal_3/en.py @@ -1,27 +1,27 @@ # coding=utf-8 name = 'pascal/3' -slug = 'pascal's triangle' +slug = 'pascal’s triangle' description = '''\ <p>The first five rows of the Pascal's triangle look like this:</p> <pre> - 1 - 1 1 - 1 2 1 - 1 3 3 1 - 1 4 6 4 1 + 1 + 1 1 + 1 2 1 + 1 3 3 1 +1 4 6 4 1 </pre> <p> Each row begins and ends with 1. Every other element can be obtained as a sum of the two values above it. Write the predicate <code>pascal(I,J,N)</code> that returns the <code>J</code>-th value in the <code>I</code>-th column of the Pascal's triangle. Your solution should return exactly one answer for any input (the <code>I</code> and <code>J</code> arguments start counting with 0; you can assume that 0 ≤ <code>J</code> ≤ <code>I</code>). <pre> - ?- pascal(0, 0, N). - N = 1. - ?- pascal(2, 1, N). - N = 2. - ?- pascal(4, 3, N). - N = 4. +?- pascal(0, 0, N). + N = 1. +?- pascal(2, 1, N). + N = 2. +?- pascal(4, 3, N). + N = 4. </pre>''' hint = {} |