Exponential digital sums: Difference between revisions
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→{{header|Wren}}: Changed to Wren S/H
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=={{header|C++}}==
{{trans|Wren}}
{{libheader|GMP}}
<syntaxhighlight lang="cpp">#include <iostream>
#include <vector>
#include <gmpxx.h>
using big_int = mpz_class;
int digit_sum(const big_int& n) {
int sum = 0;
for (char c : n.get_str())
sum += c - '0';
return sum;
}
void exp_digit_sums(int count, int min_ways, int max_power) {
for (int i = 2; count > 0; ++i) {
big_int n = i;
std::vector<int> powers;
for (int p = 2; p <= max_power; ++p) {
n *= i;
if (digit_sum(n) == i)
powers.push_back(p);
}
if (powers.size() >= min_ways) {
--count;
auto it = powers.begin();
std::cout << i << "^" << *it++;
for (; it != powers.end(); ++it)
std::cout << ", " << i << "^" << *it;
std::cout << '\n';
}
}
}
int main() {
std::cout << "First twenty-five integers that are equal to the digital sum "
"of that integer raised to some power:\n";
exp_digit_sums(25, 1, 100);
std::cout << "\nFirst thirty that satisfy that condition in three or more "
"ways:\n";
exp_digit_sums(30, 3, 500);
}</syntaxhighlight>
{{out}}
<pre>
First twenty-five integers that are equal to the digital sum of that integer raised to some power:
7^4
8^3
9^2
17^3
18^3, 18^6, 18^7
20^13
22^4
25^4
26^3
27^3, 27^7
28^4, 28^5
31^7
34^7
35^5
36^4, 36^5
40^13
43^7
45^6
46^5, 46^8
53^7
54^6, 54^8, 54^9
58^7
63^8
64^6
68^7
First thirty that satisfy that condition in three or more ways:
18^3, 18^6, 18^7
54^6, 54^8, 54^9
90^19, 90^20, 90^21, 90^22, 90^28
107^11, 107^13, 107^15
181^16, 181^18, 181^19, 181^20
360^45, 360^46, 360^49, 360^51
370^48, 370^54, 370^57, 370^59
388^32, 388^35, 388^36
523^39, 523^42, 523^44, 523^45
603^44, 603^47, 603^54
667^48, 667^54, 667^58
793^57, 793^60, 793^64
1062^72, 1062^77, 1062^81
1134^78, 1134^80, 1134^82, 1134^86
1359^92, 1359^98, 1359^102
1827^121, 1827^126, 1827^131
1828^123, 1828^127, 1828^132
2116^140, 2116^143, 2116^147
2330^213, 2330^215, 2330^229
2430^217, 2430^222, 2430^223, 2430^229, 2430^230
2557^161, 2557^166, 2557^174
2610^228, 2610^244, 2610^246
2656^170, 2656^172, 2656^176
2700^406, 2700^414, 2700^420, 2700^427
2871^177, 2871^189, 2871^190
2934^191, 2934^193, 2934^195
3077^187, 3077^193, 3077^199
3222^189, 3222^202, 3222^210
3231^203, 3231^207, 3231^209
3448^215, 3448^221, 3448^227
</pre>
=={{header|jq}}==
'''Adapted from [[#Wren|Wren]]'''
'''Works with gojq, the Go implementation of jq'''
gojq has infinite-precision integer arithmetic,
but when working on the
second task, becomes noticeably slow after producing the first ten lines or so.
<syntaxhighlight lang="jq">
# whilst/2 is like while/2 but emits the final term rather than the first one
def whilst(cond; update):
def _whilst:
if cond then update | (., _whilst) else empty end;
_whilst;
def digitSum:
def add(s): reduce s as $_ (0; . + $_);
add(tostring | explode[] | . - 48);
def expDigitSums(numNeeded; minWays; maxPower):
{i: 1, c: 0}
| whilst(.c < numNeeded;
.i += 1
| .n = .i
| reduce range(2; 1+ maxPower) as $p (.res = [];
.n *= .i
| if .i == (.n|digitSum)
then .res += ["\(.i)^\($p)"]
end)
| if .res|length >= minWays
then .c += 1
end )
| select(.res|length >= minWays)
| .res | join(", ");
"First twenty-five integers that are equal to the digital sum of that integer raised to some power:",
expDigitSums(25; 1; 100),
"\nFirst thirty that satisfy that condition in three or more ways:",
expDigitSums(30; 3; 500)
</syntaxhighlight>
{{output}}
<pre>
First twenty-five integers that are equal to the digital sum of that integer raised to some power:
7^4
8^3
9^2
17^3
18^3, 18^6, 18^7
20^13
22^4
25^4
26^3
27^3, 27^7
28^4, 28^5
31^7
34^7
35^5
36^4, 36^5
40^13
43^7
45^6
46^5, 46^8
53^7
54^6, 54^8, 54^9
58^7
63^8
64^6
68^7
First thirty that satisfy that condition in three or more ways:
18^3, 18^6, 18^7
54^6, 54^8, 54^9
90^19, 90^20, 90^21, 90^22, 90^28
107^11, 107^13, 107^15
181^16, 181^18, 181^19, 181^20
360^45, 360^46, 360^49, 360^51
370^48, 370^54, 370^57, 370^59
388^32, 388^35, 388^36
523^39, 523^42, 523^44, 523^45
603^44, 603^47, 603^54
667^48, 667^54, 667^58
793^57, 793^60, 793^64
1062^72, 1062^77, 1062^81
1134^78, 1134^80, 1134^82, 1134^86
1359^92, 1359^98, 1359^102
1827^121, 1827^126, 1827^131
1828^123, 1828^127, 1828^132
2116^140, 2116^143, 2116^147
2330^213, 2330^215, 2330^229
2430^217, 2430^222, 2430^223, 2430^229, 2430^230
2557^161, 2557^166, 2557^174
2610^228, 2610^244, 2610^246
2656^170, 2656^172, 2656^176
2700^406, 2700^414, 2700^420, 2700^427
2871^177, 2871^189, 2871^190
2934^191, 2934^193, 2934^195
3077^187, 3077^193, 3077^199
3222^189, 3222^202, 3222^210
3231^203, 3231^207, 3231^209
3448^215, 3448^221, 3448^227
</pre>
=={{header|Julia}}==
Line 68 ⟶ 279:
println("$n: powers $v")
end
end
found1 >= wanted && found2 >= multiswanted && break
end
end
Line 76 ⟶ 287:
testdigitalequals(6000, 25, 30)
</syntaxhighlight>{{out}}
<pre>
First 25 integers that are equal to the digital sum of that integer raised to some power:
Line 114 ⟶ 325:
3448: powers [215, 221, 227]
</pre>
=={{header|Phix}}==
Line 161 ⟶ 371:
<!--</syntaxhighlight>-->
Output same as Wren/Raku, second part stops on the 18th (2116) under JS but would match after a 26.5s blank screen if you used the same limits
=={{header|Python}}==
<syntaxhighlight lang="python">""" rosettacode.org/wiki/Exponential_digital_sums """
def equal_digitalsum_exponents(num):
""" return any equal exponential digital sums in an array """
equalpows = []
if num > 1:
npow, misses = 2, 0
while misses < num + 20:
dsum = sum(map(int, str(num**npow)))
if npow > 10 and dsum > 2 * num:
break # bail here for time contraints (see Wren example)
if dsum == num:
equalpows.append(npow)
else:
misses += 1
npow += 1
return equalpows
if __name__ == '__main__':
found1, found2, multis = 0, 0, []
print('First 25 integers that are equal to the digital sum of the number raised to some power:')
for i in range(1, 4000):
a = equal_digitalsum_exponents(i)
if a:
S_EXP = ', '.join(f'{i}^{j}' for j in a)
found1 += 1
if found1 <= 25:
print(S_EXP)
if len(a) > 2:
found2 += 1
multis.append(S_EXP)
if found2 == 30:
print(
'\n\nFirst 30 that satisfy that condition in three or more ways:')
for grp in multis:
print(grp)
if found1 >= 25 and found2 >= 30:
break
</syntaxhighlight>{{out}} Same as Raku.
=={{header|Raku}}==
Line 241 ⟶ 496:
Although it would be possible to use BigInt for this, GMP has been used instead to quicken up the search but even so still takes 110 seconds to run.
<syntaxhighlight lang="
var digitSum = Fn.new { |bi|
Line 344 ⟶ 599:
So, if we change the ''expDigitSums'' function to the following:
<syntaxhighlight lang="
var i = Mpz.one
var c = 0
| |||