SEND + MORE = MONEY: Difference between revisions
Added Algol 68 translation of the Julia sample
(→{{header|Wren}}: As pointed out in Julia example, S can only be 8 or 9.) |
(Added Algol 68 translation of the Julia sample) |
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Write a program in your language to solve [https://mindyourdecisions.com/blog/2018/09/06/send-more-money-a-great-puzzle/ SEND + MORE = MONEY: A Great Puzzle].
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=={{header|ALGOL 68}}==
{{Trans|Julia}}
This task <i>can</i> be solved wothout using seven nested loops but then again, it can be solved with them - so why not?.
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Uses the observations of the Julia sample (unsuprisingly as this is a translation of the Julia sample).
<syntaxhighlight lang="algol68">
BEGIN # solve the SEND+MORE=MONEY puzzle - translation of the Julia sample #
INT m = 1;
OP C = ( INT n )CHAR: REPR ( ABS "0" + n ); # convert integer to a digit #
FOR s FROM 8 TO 9 DO
FOR e FROM 0 TO 9 DO
IF e /= m AND e/= s THEN
FOR n FROM 0 TO 9 DO
IF n /= m AND n /= s AND n /= e THEN
FOR d FROM 0 TO 9 DO
IF d /= m AND d /= s AND d /= e AND d /= n THEN
FOR o FROM 0 TO 9 DO
IF o /= m AND o /= s AND o /= e AND o /= n AND o /= d THEN
FOR r FROM 0 TO 9 DO
IF r /= m AND r /= s AND r /= e AND r /= n AND r /= d AND r /= o THEN
FOR y FROM 0 TO 9 DO
IF y /= m AND y /= s AND y /= e AND y /= n AND y /= d AND y /= o AND y /= r THEN
IF ( 1000 * ( s + m ) ) + ( 100 * ( e + o ) ) + ( 10 * ( n + r ) ) + ( d + e )
= ( 10 000 * m ) + ( 1000 * o ) + ( 100 * n ) + ( 10 * e ) + y
THEN
print( ( C s, C e, C n, C d, " + ", C m, C o, C r, C e, " = ", C m, C o, C n, C e, C y ) )
FI
FI
OD
FI
OD
FI
OD
FI
OD
FI
OD
FI
OD
OD
END
</syntaxhighlight>
{{out}}
<pre>
9567 + 1085 = 10652
</pre>
=={{header|Julia}}==
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