Digit fifth powers: Difference between revisions
Added XPL0 example.
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{{Draft task}}
;Task:
<br>Task desciption is taken from Project Eulet(https://projecteuler.net/problem=30)
<br>Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
<br><br>
=={{header|Ring}}==
<lang ring>
see "working..." + nl
sumEnd = 0
sumList = []
limitStart = 10000
limitEnd = 99999
for n = limitStart to limitEnd
sum = 0
nStr = string(n)
for m = 1 to len(nStr)
sum = sum + pow(number(nStr[m]),5)
next
if sum = n
add(sumList,n)
sumEnd += n
ok
next
see "The sum of all the numbers that can be written as the sum of fifth powers of their digits:" + nl
for n = 1 to len(sumList)-1
see "" + sumList[n] + " + "
next
see "" + sumList[n] + " = " + sumEnd + nl
see "done..." + nl
</lang>
{{out}}
<pre>
working...
The sum of all the numbers that can be written as the sum of fifth powers of their digits:
54748 + 92727 + 93084 = 240559
done...
</pre>
=={{header|XPL0}}==
<lang XPL0>\upper bound: 6*9^5 = 354294
\7*9^5 is still only a 6-digit number, so 6 digits are sufficient
int A, B, C, D, E, F, \digits, A=LSD
A5, B5, C5, D5, E5, F5, \digits to 5th power
A0, B0, C0, D0, E0, F0, \digits multiplied by their decimal place
N, \number that can be written as the sum of its 5th pwrs
S; \sum of all numbers
[S:= 0;
for A:= 0, 9 do \for all digits
[A5:= A*A*A*A*A;
A0:= A;
for B:= 0, 9 do
[B5:= B*B*B*B*B;
B0:= B*10;
for C:= 0, 9 do
[C5:= C*C*C*C*C;
C0:= C*100;
for D:= 0, 9 do
[D5:= D*D*D*D*D;
D0:= D*1000;
for E:= 0, 9 do
[E5:= E*E*E*E*E;
E0:= E*10000;
for F:= 0, 3 do
[F5:= F*F*F*F*F;
F0:= F*100000;
[N:= F0 + E0 + D0 + C0 + B0 + A0;
if N = A5 + B5 + C5 + D5 + E5 + F5 then
[S:= S + N;
IntOut(0, N);
CrLf(0);
];
];
];
];
];
];
];
];
CrLf(0);
IntOut(0, S);
CrLf(0);
]</lang>
{{out}}
<pre>
0
4150
1
4151
93084
92727
54748
194979
443840
</pre>
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