Raku By Example

## ACGT

my @a=<A G C T>;
my $x=@a;
for 1 ... * -> $a {  (( [X~] $x xx $a )).join(',').say;last if $a==4;   };

## digit sum

sub digits( $_1, $_2 ) {
    return $_1, $_2, { ($^a ~ $^b) % 9 }  ...  *;
}

for 1..5 X 1..5 -> @_ {
    say digits( | @_ )[^10];
}

## 向下计数

for 10 ... 0 {
    .say;
}

## 斐波那契数列

sub fib {
    my @fib = 1, 1, *+*, ... *;
    say @fib[^20];
}

sub MAIN($name) {
    say "hello" ~ $name;
    fib();
}

## 无限序列

loop {
    say 'SPAM';
}

# In addition, there are various ways of writing lazy, infinite lists in Perl 6: 
print "SPAM\n" xx *;      # repetition operator
print "SPAM\n", ~* ... *; # sequence operator
map {say "SPAM"}, ^Inf;   # upto operator

## generate range

.say for (-270, -255 ... 0, 2 , 4 ... 10).rotor( 2 => -1);

## 100 扇门
# 问题： 你有 100 扇关着的门排成一排， 然后你穿过这些门 100 次。第一次穿过的时候，穿越每一扇门， 如果门是开着的就关闭它， 如果门是关着的就打开它。第二次穿越的时候，每两扇门穿越一下，（第 2、4、6扇门）；第三次穿越的时候， 每 3 扇门（第 3、6、9），等等， 直到你穿过第 100 扇门为止。
# 问： 最后一次穿过门之后， 每扇门的状态是开是关？
# 提示： 剩下开着的门就是那些能开方的整数the only doors that remain open are whose numbers are perfect squares of integers
#`(
my @doors = False xx 101;
($_ = !$_ for @doors[0, * + $_ ...^ * > 100]) for 1..100;
say "Door $_ is ", <closed open>[ @doors[$_] ] for 1..100;
)

say "Door $_ is open" for map {$^n ** 2}, 1..10;
say "Door $_ is open" for 1..10 X** 2;
say "Door $_ is ", <closed open>[.sqrt == .sqrt.floor] for 1..100;
Examples
