6 I can return anything over 261? kthnx
source | link

Java, 209*0200*0.6 = 125.4120

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;=s++<999;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -11000 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well. The best I could do without bonus code was 156, so it was well worth it.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;=s++<999;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

Java, 209*0.6 = 125.4

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well. The best I could do without bonus code was 156, so it was well worth it.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

Java, 200*0.6 = 120

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<999;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns 1000 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well. The best I could do without bonus code was 156, so it was well worth it.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<999;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

5 added 73 characters in body
source | link

Java, 209*0.6 = 125.4

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well. The best I could do without bonus code was 156, so it was well worth it.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

Java, 209*0.6 = 125.4

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

Java, 209*0.6 = 125.4

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well. The best I could do without bonus code was 156, so it was well worth it.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

4 snagged another bonus
source | link

Java, 171*0209*0.856 = 145125.35 bytes4

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the 0.85 bonus. Printing in Java takes way too much to bother with the other one (probably even with the updated bonus amounts, we'll see).


Below refers to the old bonus scoring onlyOld Answer: 171*0.85 = 145.35 bytes
Sadly, it takes exactly 15 bytes for me to track Lychrel candidates, so I get the exact same 156 score if I let it loop forever and claim no bonus:

import java.math.*;int f(BigInteger a){int s=0;fors=-1;for(String b,c;!(b=a+"").equals(c=new StringBuffer(b).reverse()+"");s++!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}
 

Java, 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}

With line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
            a=a.add(new BigInteger(c));
    return s>261?-1:s;
}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the 0.85 bonus. Printing in Java takes way too much to bother with the other one (probably even with the updated bonus amounts, we'll see).


Below refers to the old bonus scoring only
Sadly, it takes exactly 15 bytes for me to track Lychrel candidates, so I get the exact same 156 score if I let it loop forever and claim no bonus:

import java.math.*;int f(BigInteger a){int s=0;for(String b,c;!(b=a+"").equals(c=new StringBuffer(b).reverse()+"");s++)a=a.add(new BigInteger(c));return s;}

Java, 209*0.6 = 125.4

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));return s>261?-1:s;}

This is a simple loop that does just what it says on the box, but with some golf added. Returns -1 for Lychrel candidates to get the detection bonus. Turns out I was able to print for not too many characters (for Java at least) and snag that bonus as well.

With some line breaks:

import java.math.*;
int f(BigInteger a){
    int s=-1;
    for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)
        System.out.println(b+" + "+c+" = "+(a=a.add(new BigInteger(c))));
    return s>261?-1:s;
}

Old Answer: 171*0.85 = 145.35 bytes

import java.math.*;int f(BigInteger a){int s=-1;for(String b,c;(b=a+"").equals(c=new StringBuffer(b).reverse()+"")!=s++<262;)a=a.add(new BigInteger(c));return s>261?-1:s;}
 
3 added 4 characters in body
source | link
2 bonuses changed
source | link
1
source | link