Write a program that takes no input and outputs an integer n which is a multiple of 3. So n=3k, where k is a positive integer.
Double the last character of your code.
Your code should then output k.
Here is my program hui#2(21+4)//lm, which outputs 81.
Because of that, hui#2(21+4)//lmm should output 27.
or
If p8$w2*8 outputs 9, p8$w2*88 should output 3.
This is code-golf, shortest answer in bytes wins!
4!
4! returns 24
4!! returns 8
16-1
Outputs 15.
16-11
Outputs 5.
This is a CW answer for many languages it works for. Edit to add to the list:
16 with a constant to save a byte).
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\(@>>>
\>>%?
\(13$
\(@>>>
\>>%?
\(13$$
Lost is a language in which the start location and direction of the pointer are entirely random. Thus making a deterministic program can be tough.
The code we want to execute is either 13$(@ or 13$$(@.
31 pushes 3 and 1 to the stack$ will swap them$ will swap them back if present( will pop the top element@ will halt and outputWe want to insert the $ at the end so we have to split this
\(@
v
\31$
The rest of this is a careful orchestration to make sure that every possible pointer executes this code and only this code.
a="3";aa="1";hint a
Outputs:
a="3";aa="1";hint aa
Outputs:
hint a,aa?
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hint format["%1\n%2", a, aa] would do that (or hint (a+"\n"+aa) because they are strings), but I think outputting them separately illustrates the solution better here.
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3ȷ
Outputs 3000. 3ȷ is a numeric literal for 3000.
3ȷȷ outputs 1000: 3ȷ is 3000, which is thrown away and replaced by ȷ, 1000.
This is the only two-byte solution, and it is optimal; I found it by trying every possible program.
This prints 3 and 6ee prints 1.
6e
6ee
6 - Push 6.
e - Floor halve.
- Print Implicitly.
Not much to explain. The whole point of this solution is that:
gcd 3 21
Outputs 3, and gcd 3 211 outputs 1.
½ really square root? I can't figure which is worse, the ½ or Actually's ¬
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5Í, maybe? :P
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тg too ;)...?
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Works by changing the path based on cubes with different edge sizes. None of that tricky math stuff. The starting numbers could be also 1 3 or 2 6
39O@W@
Cubified
3
9 O @ W
@
Only executes 9O@ which is Push 9, Output as number, Halt.
39O@W@
Cubified
3 9
O @
W @ @ . . . . .
. . . . . . . .
. .
. .
This now executes W3O@, which is Left shift, Push 3, Output as number, Halt.
PrimeOmega@27
Not the shortest solution but y'know...
<q%5b'
Outputs via exit code.
Explanation: On program start the IP is immediately sent leftwards, so it wraps to the right end and executes commands from right to left. So, in this program, it encounters 'b5%, which pushes 98, the value of the character b, and then computes the remainder of 98 when divided by 5, which is 3. Finally, with q, it quits with this exit code.
With the last character doubled, it encounters ''b5%, which pushes the value of the character ' but then immediately pushes b, which is 11. It then computes the remainder of 11 when divided by 5, which is 1, and quits with this exit code.
Thanks to @nwellnhof for saving 15 bytes!
#define AA"5");}
#define A"1"AA
main(){puts(A
#define AA"5");}
#define A"1"AA
main(){puts(AA
This prints 9. x8 11 prints 3.
x8 1
x8 11
x8 1 - Full program.
x - Bitwise XOR.
8 1 - Of 8 and 1.
- Implicitly print.
[q/3g
Outputs via exit code.
First, the IP hits the "turn left" instruction ([) twice, which wraps it back around to the back of the code, going right.
The "get" instruction (g) takes the first 2 values and pushes the ASCII value of the character at that x and y position. The first time it is executed, there are only zeros on the stack, meaning we get the value 91. If we don't double the g, we divide 91 by 3, with integer division, truncating it to 30, which is returned.
If we do duplicate the g, it gets the ASCII value at (0, 91), which isn't anywhere near the code. Because the grid is initialized to be filled with spaces, this returns the value 32. if we integer divide by 3, we get 10, one third of 30.
a=3;aa=1;print a
outputs 3
a=3;aa=1;print aa
outputs 1
+16-1
- sign and subtract operator are the same. This is actually adding 16 and -1.Y_+B5A3IH1
Still on the hunt for a 2 byte solution.
#Q is 81 and ¬ is square root, which gives 9. Adding another ¬ at the end gives 3
#Q¬
#Q¬¬
",n;c!
",n;c!!
",n;c! : Put each character onto the stack;
[110, 59, 99, 33]
,n; : Divide the top two stack items, print as a number and end.
99/33 = 3
",n;c!! : Put each chat onto the stack;
[110, 59, 99, 33, 33]
,n; : Divide the top two stack items, print as a number and end.
99/33 = 3
9.0and3.0? (Taking square roots, so it automatically turns into a float. :P) \$\endgroup\$