Leave your mark!

Question

In this challenge, you'll need to edit a grid of characters to make a program that does a task, while ensuring that you leave a lasting mark on the grid.

Task

The task to perform is determined by when you answer. Specifically, for the nth answer, divide n by 5, and if the remainder is...

• 0. Read a positive integer from STDIN (or closest alternative) and print its square.
• 1. Print the value of n⁹-n⁸+n⁷-...-n²+n-1.
• 2. Read a string from STDIN (or closest alternative) and print it, reversed.
• 3. Print the name of the language being used.
• 4. Print the first 99 Fibonacci numbers, with some delimiter.

In all cases, "print" means to STDOUT or closest alternative. Leading or trailing whitespace is okay. A full program is required, not just a function.

Answering (please read before posting)

This is a "sequential answering" challenge. Answers have a specific order. If the previous answer, chronologically, was titled Answer n, your answer should be titled Answer n+1.

All answers are 100 characters (in printable ASCII) long, divided into ten rows. Each row is separated by a newline. I have already posted the first answer.

Once an answer has been verified by another user (in a comment), any user, except for the one who gave the previous answer, can submit the next answer, pursuant to the following rules:

For each answer after the first, change at least one character (but not newlines) in the previous answer to another printable ASCII character achieve the correct task. You may not add, remove, or move any characters. You may use any language with a readily available compiler or interpreter.

Furthermore, you must specify one character you changed to be your "mark". You want to maximize the longevity of your mark. If your mark is first changed by the answer five after yours, its lifespan is 4 (four other answers left it untouched). If no one ever removes your mark, then the lifespan is the total number of other answers after it.

If you answer multiple times, you may have multiple existing marks at once.

If your answer is found incorrect, please delete it if there are no subsequent valid answers. Otherwise, do not delete or change your answers.

Scoring

Your score is the sum of your marks' longevities, divided by the total number of characters you changed. Subtract 5% from your score for each answer past the first.

For example, suppose I answer three times. First with 5 changes and a mark longevity of 3, then 4 changes and 20, then 1 change and a longevity of 7. My score is now ((3 + 20 + 7) / (5 + 4 + 1)) * (100% - 10%) = 2.7.

Answer 1

Answer 2, CJam, 6 changed

"lass Fun{
 public   
  static  
void main(
 String[] 
args) {   
System.out
.print(0);
  }       
}";lW%    

The mark is the ; in the last line.

Test it here.

Answer 2

Answer 5, Deadfish, changes 4

Deadfish has no input, thus the closest alternative to STDIN is that the number is already stored in the counter. Note that the language can't handle numbers over 255, so the "input" must be between 0 and 15 inclusive.

The j in statjc is the mark.

 var  Fun,
 publjc,  
  statjc, 
v=2;  f3r(
 S =[1,1];
v<99;){   
S[v]=S[v-1
]+S[v++-2]
  o       
alert(S)  

Answer 3

Answer 3, CJam, 8 changed

"lass Fun{
 public   
  static  
void main(
 String[] 
args) {   
System.out
.print(0);
  }       
";"CJam"  

Changed the mark (;) from Answer 2.

My mark is the second " on the last line.

Test it here.

Answer 4

Answer 6, CJam, changes 13

"var  Fun,
 publjc,  
";8638UY5"
v=2;  f3r(
 S =[1,1];
v<99;){   
S[v]=S[v-1
]+S[v++-2]
  o       
alert(S)";

Try it online here

My mark is the last ;

Answer 5

Answer 4, JavaScript, 56 changed

 var  Fun,
 public,  
  static, 
v=2;  for(
 S =[1,1];
v<99;){   
S[v]=S[v-1
]+S[v++-2]
  }       
alert(S)  

Removed the previous mark. My mark is the } in the second-to-last line. Unfortunately, JavaScript rounds the really big FIbonacci numbers, but I think it works. (Note that there should be a leading space and 2 trailing spaces that are automatically removed.)

Answer 6

Answer 7, ><>, changes 14

i:0(  !#?^
 publjc,  
";8638UY5o
v=2;  f3r:
 S =[1,1];
v<99;){  !
S[v]=S[v-?
]+S[v++-2l
  o       
alert(S)"~

My mark is the # on the first line.

Answer 7

Answer 1, Java

class Fun{
 public   
  static  
void main(
 String[] 
args) {   
System.out
.print(0);
  }       
}