Munchausen numbers: Difference between revisions
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Alternative that finds all 4 Munchausen numbers. As noted by the Pascal sample, we only need to consider one arrangement of the digits of each number (e.g. we only need to consider 3345, not 3435, 3453, etc.). This also relies on the non-standard 0^0 = 0.
<lang algol68># Find all Munchausen numbers - note 11*(9^9) has
# Munchausen numbers with 11+ digits #
# table of Nth powers - note 0^0 is 0 for Munchausen numbers, not 1 #
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[ 0 : 9 ]INT d count := z count;
# as the digit
# consider one arrangement of each possible combination of digits #
FOR d1 FROM 0 TO 9 DO
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digit power sum +:= nth power[ d7 ] + nth power[ d8 ];
digit power sum +:= nth power[ d9 ] + nth power[ da ];
# count the occurrences of each digit (including leading zeros #▼
d count := z count;
d count[ d1 ] +:= 1; d count[ d2 ] +:= 1; d count[ d3 ] +:= 1;
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d count[ d7 ] +:= 1; d count[ d8 ] +:= 1; d count[ d9 ] +:= 1;
d count[ da ] +:= 1;
# subtract the occurrences of each digit in the power sum #▼
# (also including leading zeros) - if all counts drop to 0 we #▼
LONG INT number := digit power sum;
d count[ SHORTEN ( number MOD 10 ) ] -:= 1;
number OVERAB 10
OD;
IF d count[ 0 ] = 0 AND d count[ 1 ] = 0 AND d count[ 2 ] = 0
AND d count[ 3 ] = 0 AND d count[ 4 ] = 0 AND d count[ 5 ] = 0
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