Dinesman's multiple-dwelling problem: Difference between revisions
Dinesman's multiple-dwelling problem (view source)
Revision as of 06:31, 15 July 2023
, 11 months agorust example
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<pre>
"Smith Cooper Baker Fletcher Miller"
</pre>
=={{header|Rust}}==
<syntaxhighlight lang = "rust">use itertools::Itertools;
fn main() {
for p in (1..6).permutations(5) {
let baker: i32 = p[0];
let cooper: i32 = p[1];
let fletcher: i32 = p[2];
let miller: i32 = p[3];
let smith: i32 = p[4];
if baker != 5 && cooper != 1 && fletcher != 1 && fletcher != 5 && cooper < miller &&
(smith - fletcher).abs() > 1 && (cooper - fletcher).abs() > 1 {
print!("Baker on {baker}, Cooper on {cooper}, ");
println!("Fletcher on {fletcher}, Miller on {miller}, Smith on {smith}.");
break;
}
}
}
</syntaxhighlight>{{out}}
<pre>
Baker on 3, Cooper on 2, Fletcher on 4, Miller on 5, Smith on 1.
</pre>
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