Sum multiples of 3 and 5: Difference between revisions
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m (→{{header|MiniScript}}: verified and corrected program(s) output.) |
(Updated first program to compile with Nim 1.4. Added general comment. Added output for first example. Updated output for second example.) |
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=={{header|Nim}}== |
=={{header|Nim}}== |
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Here is the solution using normal integers. |
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<lang nim>proc sum35(n: int): int = |
<lang nim>proc sum35(n: int): int = |
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for x in 0 .. |
for x in 0 ..< n: |
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if x mod 3 == 0 or x mod 5 == 0: |
if x mod 3 == 0 or x mod 5 == 0: |
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result += x |
result += x |
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echo sum35(1000)</lang> |
echo sum35(1000)</lang> |
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{{out}} |
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With BigInts: |
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<pre>233168</pre> |
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To compute until 1e20, we have to use big integers. As Nim doesn’t provided them in its library, we have to use a third party library, either "bigints" or "bignum". |
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{{trans|Raku}} |
{{trans|Raku}} |
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{{libheader|bigints}} |
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<lang nim>import bigints |
<lang nim>import bigints |
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echo sum35 x |
echo sum35 x |
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x *= 10</lang> |
x *= 10</lang> |
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Output: |
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{{out}} |
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<pre> |
<pre>0 |
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23 |
23 |
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2318 |
2318 |
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