本文主要是介绍程序员的算法趣题Perl版（四），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
间隔的时间有点长，但是我并没有放弃啊同志们~~~好吧，只有我自己看。。。
第十一题
用斐波那契数列中的每一个数除以其数位上所有数字之和，求出能整除的数。我这里设置的范围是前1000个数字。
#! perl# 20180108# this program is used to find out some numbers in Fibonacci , and the numbers can be divisible
# by the sum of its every single number;
use strict;# the special Fibonacci;
my @fibonacci;
# the maximum of times that the recusive function is executed
my $max = 1000;
&createFibonacci(1, 1, 0, \@fibonacci);# print
foreach(@fibonacci) {print $_ . "\n";
}sub createFibonacci {my $first = shift;my $second = shift;# the my $count = shift;# the array pointmy $p = shift;my $third = $first + $second;# push if the number is matchedpush @$p, $third if &check($third);$count ++;# the length of array now;my $size = @$p;return 0 if($size == 11);return 0 if $count == $max;# if not contiue...&createFibonacci($second, $third, $count, $p);
}sub check {my $num = shift;my $leng = length $num;my $sum = &sum($num, $leng);return 1 if ($num % $sum) == 0;0;
}# this function is used to sum a number`s every single numbers;
sub sum {my $num = shift;my $leng = shift;my $sum = 0;for(my $i = 0; $i < $leng; $i ++) {return $num if $leng == 1;$sum += substr $num, $i, 1;}$sum;
}答案：
2
3
5
8
21
144
2584
14930352
86267571272
498454011879264
160500643816367088
第十二题
求在计算平方根的时候，最早让0-9全部出现的最小整数。
#! perl# 20180109
# 20180110use strict;for(my $i = 2; $i < 5000; $i ++) {if(&check($i, 0)) {print "$i\n";last;}
}sub check {# the number checkedmy $num = shift;# '0' means that includes integer, '1' means just decimalmy $justDecimal = shift;# get the square root of a numbermy $sqrtR = sprintf "%10.10f", sqrt $num;# split with '.'my @allNum = split /\./, $sqrtR;# this array is used to check if the number is repeatmy @arr;if($justDecimal) {# just split the decimal part@arr = split //, $allNum[1];} else {my @integerArr = split //, $allNum[0];my @decimalArr = split //, $allNum[1];# include integer, get 10 numbers from the left@arr = (@integerArr, @decimalArr)[0..9];}&isRepeat(\@arr);
}sub isRepeat {# the point of the arraymy $p = shift;# key is numbers, value is timesmy %hash = ();my @uniqueArr = grep {# put the number which is found only once into the hash++ $hash{$_} < 2;} @$p;# the array is repeat if the size of the array is not 10return 1 if $#uniqueArr + 1 == 10;
}答案：
包含整数部分：1362
只有小数部分：143
第十三题
给字母赋值，求使下列算式成立的解有多少种。字母代表的数字不可以相等。
#! perl# 20180114
# 20180116# this guy is used to compute READ + WRITE + TALK = SKILL;use Algorithm::Combinatorics qw(combinationscombinations_with_repetitionvariationsvariations_with_repetitiontuplestuples_with_repetitionpermutationscircular_permutationsderangementscomplete_permutationspartitionssubsets);
use strict;my $start = time;
my @numbers = (0..9);my @result;
my $count = 0;&sum(\@numbers, 10);foreach (@result) {print "$_\n";
}print "$count \n";my $duration = time - $start;
print "$duration s\n";sub sum {my $numbersP = shift;my $k = shift;# permutationmy $iter = variations($numbersP, $k);while(my $item = $iter -> next) {&isRight($item);}
}sub isRight {my @str_arr = @{@_[0]};my ($r, $e, $a, $d, $w, $i, $t, $l, $k, $s) = @str_arr;# The highest number of digits can not be zero.return 0 if $r == 0 || $w == 0 || $t == 0 || $s == 0;my $read = $r * 1000 + $e * 100 + $a * 10 + $d;my $write = $w * 10000 + $r * 1000 + $i * 100 + $t * 10 + $e;my $talk = $t * 1000 + $a * 100 + $l * 10 + $k;my $skill = $s * 10000 + $k * 1000 + $i * 100 + $l * 10 + $l;if ($read + $write + $talk == $skill) {push @result, "$read + $write + $talk = $skill";$count ++;return 1;}0
}答案：
10种
1632 + 41976 + 7380 = 50988
2543 + 72065 + 6491 = 81099
4905 + 24689 + 8017 = 37611
5094 + 75310 + 1962 = 82366
5096 + 35710 + 1982 = 42788
5180 + 65921 + 2843 = 73944
5270 + 85132 + 3764 = 94166
7092 + 37510 + 1986 = 46588
7092 + 47310 + 1986 = 56388
9728 + 19467 + 6205 = 35400
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