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The leaders of the world have met and have finally admitted that the best (and only) way of resolving global economic woes is to take stock of how much they owe each other and just pay each other off with huge cheques. They have hired you (ironically, at the lowest contract rate possible) to work out the best means of doing so.

After much deliberation, and asking someone to draw a simple example they've come up with the following spec.

Each country is represented by their ISO 3166-1 alpha-2 code: US for the USA, AU for Australia, JP for Japan, CN for China and so on...

  1. A ledger is drawn up as a series of country entries and the amounts owing to each country.
  2. Each country's entry starts off with their domain ID a colon, and how much they have in surplus/deficit (in billions of Euro), followed by a semicolon, then a coloned comma-separated list of the countries and how much (in billions of Euro) they owe.
  3. If a country owes nothing to another country, no mention of that country is entered after that semicolon separator.
  4. Deficits are indicated as negative numbers, surplus is indicated as a positive number.
  5. Values can also be floats.
  6. The ledger must be taken from STDIN. The end of the ledger is indicated by a carriage return on a blank line. The tally must be delivered to STDOUT.

An example of a ledger:

Input:
AU:8;US:10,CN:15,JP:3
US:14;AU:12,CN:27,JP:14
CN:12;AU:8,US:17,JP:4
JP:10;AU:6,US:7,CN:10

The system then works out how much each country owes and is owed and determines their surplus/deficit, for example, for AU:

AU = 8 (current surplus) -10 (to US) -15 (to CN) -3 (to JP) +12 (from US) +8 (from CN) +6 (from JP) = 6

When all the computing is done, a tally must be shown:

Output:
AU:6
US:-5
CN:35
JP:8

Your job is to create this system, capable of taking any number of ledger entries for any number of countries and capable of determining how much each country has in deficit/surplus when everything is paid out.

The ultimate test is for you to use your code to resolve the debt owed between the following countries in the test case below. These figures were taken from BBC News as of June 2011. (http://www.bbc.com/news/business-15748696)

For the purposes of the exercise, I have used their respective GDP as their current surplus... Please bear in mind that this is strictly an exercise in code quality assurance... there will be no talk of global economic resolution here in this question... If you want to talk economics I'm sure there's another subdomain in SE that handles it...

US:10800;FR:440.2,ES:170.5,JP:835.2,DE:414.5,UK:834.5
FR:1800;IT:37.6,JP:79.8,DE:123.5,UK:227,US:202.1
ES:700;PT:19.7,IT:22.3,JP:20,DE:131.7,UK:74.9,US:49.6,FR:112
PT:200;IT:2.9,DE:26.6,UK:18.9,US:3.9,FR:19.1,ES:65.7
IT:1200;JP:32.8,DE:120,UK:54.7,US:34.8,FR:309,ES:29.5
IE:200;JP:15.4,DE:82,UK:104.5,US:39.8,FR:23.8
GR:200;DE:15.9,UK:9.4,US:6.2,FR:41.4,PT:7.5,IT:2.8
JP:4100;DE:42.5,UK:101.8,US:244.8,FR:107.7
DE:2400;UK:141.1,US:174.4,FR:205.8,IT:202.7,JP:108.3
UK:1700;US:578.6,FR:209.9,ES:316.6,IE:113.5,JP:122.7,DE:379.3

Now, be the economic savior of the world!

Rules:

  1. Shortest code wins... this is code-golf after all...
  2. Please provide your output of the major test case with your code answer...
share|improve this question
1  
In the "ultimate test", shouldn't there be a semicolon after JP:4100? –  Mathieu Rodic Mar 27 at 8:53
6  
I can't help but wonder if this is a very clever way of having a homework assignment completed for you. If so, you deserve it. –  mkingston Mar 27 at 11:30
2  
Yeah, if you did this will real numbers you'd notice an astonishing contradiction. The sum of all surpluses and deficits will be negative. –  Cruncher Mar 27 at 12:57
3  
Actually, it's not homework in disguise... It was inspired from my weekly poker tournament with my friends... Trying to figure out a quicker way of working out the winnings for each player ;) –  WallyWest Mar 27 at 20:31
1  
@WallyWest LOL ;) btw, apologies for the OT comments, but it's a matter I'm quite passionate about. Now let's get back to some fun coding and forget the woes of the world... –  Tobia Mar 27 at 20:33

15 Answers 15

K, 66

{(((!)."SF"$+":"\:'*+a)-+/'d)+/d:"F"$(!).'"S:,"0:/:last'a:";"\:'x}

.

k)input:0:`:ledg.txt
k){(((!)."SF"$+":"\:'*+a)-+/'d)+/d:"F"$(!).'"S:,"0:/:last'a:";"\:'x} input
US| 9439.3
FR| 2598.9
ES| 852.1
PT| 90.1
IT| 887.5
IE| 48
GR| 116.8
JP| 4817.4
DE| 2903.7
UK| 1546.2
share|improve this answer
    
Incredibly impressed with this one... any chance you can provide a link to the K programming paradigm? –  WallyWest Mar 31 at 22:25
    
@WallyWest code.kx.com provides lots of information on q, which is the syntactic sugar that sits on top of k. q, moreso thank k, is what you'll find in production systems but for golfing k has the edge. Also check out Kona (github.com/kevinlawler/kona) which is an open source implementation of an older version of k –  tmartin Apr 1 at 12:05

Perl, 139 137 134 119 112

Here's another working piece of code... I will document it later.

Golfed code

With dictionary (112):

for(<>){~/:(.+);/g;$d{$c=$`}+=$1;$l=$';$d{$1}+=$2,$d{$c}-=$2while$l=~/(..):([^,]+)/g}print"$_:$d{$_}
"for keys%d

Without dictionary (137):

for($T=$t.=$_ for<>;$t=~/(..:)(.+);(.+)/g;print"$c$s\n"){$c=$1;$l=$3;$s=$2;$s-=$&while$l=~/[\d.]+/g;$s+=$1while$T=~/$c([\d.]+)(?!;|\d)/g}

Output

US:9439.3
FR:2598.9
ES:852.1
PT:90.1
IT:887.5
IE:48
GR:116.8
JP:4817.4
DE:2903.7
UK:1546.2

See it in action!

http://ideone.com/4iwyEP

share|improve this answer
4  
The definition of "short" should be judged by amount of tokens, not characters. Readability 4 life! –  Domi Mar 28 at 6:57
7  
@Domi - you're new here, aren't you ;-) –  jimbobmcgee Mar 28 at 12:27
4  
@jimbobmcgee: I also have a feeling this website is not much about readability... –  Mathieu Rodic Mar 28 at 12:28

Python, 211 185 183

import sys,re;t,R,F=sys.stdin.read(),re.findall,float;S=lambda e,s:sum(map(F,R(e,s)))
for m in R('(..:)(.+);(.+)',t):print m[0]+`F(m[1])+S(m[0]+'([\d.]+)(?!;|\d)',t)-S('[\d.]+',m[2])`

Output with major test case:

US:9439.300000000001
FR:2598.9
ES:852.0999999999999
PT:90.09999999999997
IT:887.5
IE:48.0
GR:116.8
JP:4817.4
DE:2903.7
UK:1546.2000000000003

(test it here: http://ideone.com/CjWG7v)

share|improve this answer

C - 257 253 if no CR at end of line

Depends on sizeof(short)==2.

No check for buffer overflow.

#define C(c) x[*(short*)c]
main(i){double x[23131]={0},d;char*q,b[99],*(*s)()=strtok;for(;gets(b);)for(s(b,":"),C(b)+=atof(s(0,";"));q=s(0,":");C(b)-=d=(atof(s(0,","))),C(q)+=d);for(i=b[2]=0;i<23131;memcpy(b,&i,2),x[i]?printf("%s:%f\n",b,x[i++]):++i);}

Output:

DE:2903.700000  
IE:48.000000    
UK:1546.200000  
JP:4817.400000  
FR:2598.900000  
GR:116.800000   
ES:852.100000   
US:9439.300000  
IT:887.500000   
PT:90.100000   

Less golfed:

#define C(c) x[*(short*)c]

main(i)
{
    double x[23131]={0}, d;
    char *q, b[99], *(*s)()=strtok;
    for(;gets(b);) 
        for(s(b, ":"),C(b)+=atof(s(0, ";")); 
            q=s(0, ":"); 
            C(b)-=d=(atof(s(0, ","))), C(q)+=d) ;

    for(i=b[2]=0; 
        i<23131; 
        memcpy(b, &i, 2), x[i]?printf("%s:%f\n", b, x[i++]):++i) ;
}
share|improve this answer

JavaScript (ES6): 254, 248, 214, 204 199

C={};s='split';prompt()[s](/\s/).map(a=>{c=a[s](';');n=c[0][s](':');m=n[0];C[m]|=0;C[m]+=+n[1];c[1][s](',').map(b=>{b=b[s](':');e=b[0];C[e]|=0;C[m]-=+b[1];C[e]+=+b[1]})});for(x in C)alert(x+':'+C[x])

Ungolfed:

C={};
s = 'split';
prompt()[s](/\s/).map(a=>{
    c = a[s](';');
    n = c[0][s](':');
    m = n[0];

    C[m]|=0;
    C[m] += +n[1];

    c[1][s](',').map(b=>{
        b = b[s](':');
        e = b[0];

        C[e]|=0;

        C[m]-=+b[1];
        C[e]+=+b[1];
    });
});

for(x in C) alert(x+':'+C[x])

Pretty straightforward. It's probably not going to win (m)any prizes. This is my first time using the new features of ES6 (though only the arrow-operator so far). I wouldn't be surprised if there are more of those that can improve the par for this code.

I'm splitting on white-space instead of just newlines because Firefox (unlike Chrome) turns newlines in a prompt-box into spaces.

Output:

US: 9439.299999999997
FR: 2598.8999999999996
ES: 852.1
JP: 4817.4
DE: 2903.7
UK: 1546.2000000000003
IT: 887.5
PT: 90.09999999999998
IE: 48
GR: 116.8

If this script is correct then in the end no country (using the BBC's data) is in the red when all debts are settled. Leaving me once more dissapointed in economics i.e. why don't they just fix it right now instead of forcing every country into budget cuts due to their national debt. But that's probably too simplistic.

See the comments for a more accurate description of the actual problem.

Thanks to @toothbrush for the C[m]|=0; shortening. Thanks to @user2428118 for suggesting the s = 'split' trick.

share|improve this answer
    
Looks great, can you please provide the output in your answer? –  WallyWest Mar 26 at 23:59
2  
@Comintern the first number is neither a debt nor a surplus, it's the GDP, so this entire exercise of subtracting debt from it makes no economic sense. In any case, the foreign debt is owed by specific banks, therefore in the end by specific individuals, towards other countries. You can't just cancel the debt the US owes to Alice (from the UK) with the "opposite" debt the UK owes to Bob (from the US). In this example, Alice and Bob are extremely rich bankers and the people of the US and UK collectively owe them money. –  Tobia Mar 27 at 7:33
2  
You can shorten C[e]=C[e]||0 to C[e]|=0, and likewise for C[m]. –  toothbrush Mar 27 at 14:42
2  
Also, you can shorten C[e]|=C[e]||0;C[m]-=+b[1];C[e]+=+b[1] to C[e]=C[e]|0+ +b[1];C[m]-=+b[1]. –  toothbrush Mar 27 at 14:44
1  
You call split multiple times. No need to write it out in full each time: p='split',C={};prompt()[p](/\s/).map(a=>{c=a[p](';');n=c[0][p](':');m=n[0];C[m]‌​|=0;C[m]+=+n[1];c[1][p](',').map(b=>{b=b[p](':');e=b[0];C[e]|=0;C[m]-=+b[1];C[e]+‌​=+b[1]})});for(x in C)alert(x+':'+C[x]) –  user2428118 Mar 28 at 13:57

PHP - 338, 280

Should work with any version of PHP 5.

Golfed:

while(preg_match("#(..):(.+);(.*)#",fgets(STDIN),$m)){$l[$m[1]][0]=(float)$m[2];foreach(explode(",",$m[3])as$x){$_=explode(":",$x);$l[$m[1]][1][$_[0]]=(float)$_[1];}}foreach($l as$c=>$d)foreach($d[1]as$_=>$o){$l[$_][0]+=$o;$l[$c][0]-=$o;}foreach($l as$c=>$d)echo$c,":",$d[0],"\n";

Un-golfed:

<?php

while( preg_match( "#(..):(\d+);(.*)#", fgets( STDIN ), $m ) )
{
    $l[$m[1]][0] = (float)$m[2];

    foreach( explode( ",", $m[3] ) as $x )
    {
        $_ = explode( ":", $x );
        $l[$m[1]][1][$_[0]] = (float)$_[1];
    }
}

foreach( $l as $c => $d )
    foreach( $d[1] as $_ => $o )
    {
        $l[$_][0] += $o;
        $l[$c][0] -= $o;
    }

foreach( $l as $c => $d )
    echo $c, ":", $d[0], "\n";

Output:

US:9439.3
FR:2598.9
ES:852.1
PT:90.1
IT:887.5
IE:48
GR:116.8
JP:4817.4
DE:2903.7
UK:1546.2
share|improve this answer
    
Wouldn't it be shorter if you used preg_match_all() and then just looped once? –  DKasipovic Mar 28 at 10:45

perl (184 characters)

Code

%c,%d,%e=();while(<>){$_=~/(..):(.+);(.*)/;$n=$1;$c{$1}=$2;for $i(split /,/,$3){$i=~/(..):(.+)/;$d{$1}+=$2;$e{$n}+=$2;}}for $i(keys %c){$c{$i}+=$d{$i}-$e{$i};print $i.":".$c{$i}."\n";}

Output

UK:1546.2
DE:2903.7
IT:887.5
FR:2598.9
PT:90.1
US:9439.3
JP:4817.4
ES:852.1
IE:48
GR:116.8
share|improve this answer

Perl - 116 114 112

for(<>){($n,$m,@l)=split/[:;,]/;$h{$n}+=$m;$h{$n}-=$p,$h{$o}+=$p while($o,$p,@l)=@l}print"$_:$h{$_}\n"for keys%h

Output:

GR:116.8
UK:1546.2
DE:2903.7
IE:48
IT:887.5
US:9439.3
PT:90.1
ES:852.1
FR:2598.9
JP:4817.4

Ungolfed:

for(<>) {
    ($n, $m, @l)=split(/[:;,]/);
    $h{$n}+=$m;

    $h{$n}-=$p, $h{$o}+=$p while ($o,$p,@l)=@l
}
print "$_:$h{$_}\n" for keys%h
share|improve this answer
    
Nice! I like your approach :) –  Mathieu Rodic Mar 28 at 12:30

C++ - 1254

#include<iostream>
#include<cstring>
#include<vector>
#include<sstream>
#include<cstdlib>
using namespace std;int main(){vector<string>input,countries,output;vector<double>results;string last_val;int j,k,i=0;cout<<"Input\n";do{getline(cin,last_val);if(last_val!=""){input.push_back(last_val);countries.push_back(last_val.substr(0,2));}}while(last_val!="");for(j=0;j<countries.size();j++){results.push_back(0);for(k=0;k<input.size();k++)input[k].substr(0, 2)==countries[j]?results[j]+=atof(input[k].substr((input[k].find(countries[j])+3),(input[k].find(',',input[k].find(countries[j]))-input[k].find(countries[j]))).c_str()):results[j]+=atof(input[k].substr((input[k].find(countries[j],3)+3),(input[k].find(',',input[k].find(countries[j]))-input[k].find(countries[j]))).c_str());}for(j=0;j<input.size();j++){for(k=0;k<countries.size();k++){if(input[j].substr(0,2)!=countries[k]){results[j]-=atof(input[j].substr((input[j].find(countries[k])+ 3),(input[j].find(',',input[k].find(countries[k]))-input[j].find(countries[j]))).c_str());}}}for(i=0;i<countries.size();i++){stringstream strstream;strstream<<countries[i]<<":"<<results[i];output.push_back(strstream.str().c_str());}cout<<"Output:\n";for(i=0;i<output.size();i++){cout<<output[i]<<'\n';}return 0;}

I realize the code is very long, but enjoyed the good fun. This is my first time code golfing, and I am new to C++, so suggestions for improving my code are much appreciated.

Final Challenge Results

Output:
US:9439.3
FR:2598.9
ES:852.1
PT:90.1
IT:887.5
IE:48
GR:116.8
JP:4817.4
DE:2903.7
UK:1546.2

Ungolfed Code

#include<iostream>
#include<cstring>
#include<vector>
#include<sstream>
#include<cstdlib>

using namespace std;

int main() {
  vector<string> input, countries, output;
  vector<double> results;
  string last_val;
  int i, j, k;

  cout << "Input\n";
  do {
    getline(cin, last_val);
    if(last_val != "") {
      input.push_back(last_val);
      countries.push_back(last_val.substr(0, 2));
    }
  } while(last_val != "");

  for(j = 0; j < countries.size(); j++) {
    results.push_back(0);
    for(k = 0; k < input.size(); k++) {
      if(input[k].substr(0, 2) == countries[j]) {
        results[j] += atof(input[k].substr((input[k].find(countries[j]) + 3),
                             (input[k].find(',', input[k].find(countries[j])) -
                              input[k].find(countries[j]))).c_str());
      } else {
        results[j] += atof(input[k].substr((input[k].find(countries[j], 3) + 3),
                             (input[k].find(',', input[k].find(countries[j])) -
                              input[k].find(countries[j]))).c_str());
      }
    }
  }

  for(j = 0; j < input.size(); j++) {
    for(k = 0; k < countries.size(); k++) {
      if(input[j].substr(0, 2) != countries[k]) {
        results[j] -= atof(input[j].substr((input[j].find(countries[k]) + 3),
                             (input[j].find(',', input[k].find(countries[k])) -
                              input[j].find(countries[j]))).c_str());
      }
    }
  }

  for(i = 0; i < countries.size(); i++) {
    stringstream strstream;
    strstream << countries[i] << ":" << results[i];
    output.push_back(strstream.str().c_str());
  }

  cout << "Output:\n";
  for(i = 0; i < output.size(); i++) {
    cout << output[i] << '\n';
  }

  return 0;
}
share|improve this answer
1  
Hi, Nice to see an example in C++. You could reduce number of characters by using one-letter identifiers in stead of descriptive names, i.e. use i for input, c for countries and so on. –  ahy1 Mar 31 at 18:11
    
Agreed with @ahy1 here... If you reduce your variables to 1 letter you can cut a fair bit of this... You may also find this of interest for future golf challenges: codegolf.stackexchange.com/questions/132/tips-for-golfing-in-c –  WallyWest Mar 31 at 22:28
    
Oh, and you won't need cout << "Output:\n"; either... That's a 20 byte saving there... –  WallyWest Mar 31 at 22:29

AWK - 138 120

{l=split($0,h,"[:,;]");t[h[1]]+=h[2];for(i=3;i<l;i+=2){t[h[1]]-=h[i+1];t[h[i]]+=h[i+1]}}END{for(v in t){print v":"t[v]}}

And the results

$ cat data.withoutInputHeadline |awk -f codegolf.awk
IT:887.5
UK:1546.2
DE:2903.7
PT:90.1
ES:852.1
FR:2598.9
GR:116.8
Input:0
JP:4817.4
IE:48
US:9439.3

Ungolfed

{
    l=split($0,h,"[:,;]");
    t[h[1]]+=h[2];
    for(i=3;i<l;i+=2){
        t[h[1]]-=h[i+1]
        t[h[i]]+=h[i+1]
    }
}
END{
    for(v in t){
        print v":"t[v]
    }
}

(test it here: http://ideone.com/pxqc07)

share|improve this answer
    
Why even put those headers in? You'll save more bytes without them... they weren't even part of the spec I set... ;) –  WallyWest Mar 31 at 12:03
    
@WallyWest: Ok, so I didn't understood that, because they are displayed in your first example of Input and Output e.g.: (..)a tally must be shown: Output: (..) No worry about that, I remove my first example right now. –  Doomsday Mar 31 at 12:33

Ruby - 225

First try in a challenge like this, sure it could be a lot better...

R=Hash.new(0)
def pd(s,o=nil);s.split(':').tap{|c,a|R[c]+=a.to_f;o&&R[o]-=a.to_f};end
STDIN.read.split("\n").each{|l|c,d=l.split(';');pd(c);d.split(',').each{|s|pd(s,c.split(':')[0])}}
puts R.map{|k,v|"#{k}: #{v}"}.join("\n")

And the results

$ cat data|ruby codegolf.rb
US: 9439.299999999997
FR: 2598.8999999999996
ES: 852.1
JP: 4817.4
DE: 2903.7
UK: 1546.2000000000003
IT: 887.5
PT: 90.09999999999998
IE: 48.0
GR: 116.8
share|improve this answer

JS, 254 240 245

z='replace';r={};p=eval(('[{'+prompt()+'}]')[z](/\n/g,'},{')[z](/;/g,','));for(i in p){l=p[i];c=0;for(k in l){if(!c){c=k;r[c]=0;}else{r[c]-=l[k];}};for(j in p){w=p[j][c];if(w!=null)r[c]+=w}};alert(JSON.stringify(r)[z](/"|{|}/g,'')[z](/,/g,'\n'))

Well..I know it is quite long but this is my second code golf.

Suggestions are welcome!

BTW, Interesting Javascript preserves the order of elements in hashmaps, so, even if p contains an array of dictionaries, I can iterate each dictionary as an array and I'm sure that the first element of a dict is the first inserted. (the name of the country referred to the current line)

Ungolfed:

z='replace';
r={};
p=eval(('[{'+prompt()+'}]')[z](/\n/g,'},{')[z](/;/g,',')); // make the string JSONable and then evaluate it in a structure
for(i in p){ 
    l=p[i];
    c=0;
    for(k in l){
            if(!c){ // if c is not still defined, this is the country we are parsing.
                    c=k;
                    r[c]=0;
            }
            else r[c]-=l[k];
    }; 
    for(j in p){
            w=p[j][c];
            if(!w)  r[c]+=w
    }
};
alert(JSON.stringify(r)[z](/"|{|}/g,'')[z](/,/g,'\n')) # Stringify the structure, makes it new-line separated.

Note: the input is a prompt() which should be a single line. But if you copy/paste a multi line text (like the proposed input) in a prompt() window then JS read it all.

Output:

US:9439.3
FR:2598.9
ES:852.1
PT:90.09999999999998
IT:887.5
IE:48
GR:116.8
JP:4817.4
DE:2903.7000000000003
UK:1546.2
share|improve this answer
1  
You use the word "replace" four times in your code. How about shortening it like this: z='replace';r={};p=eval(('[{'+prompt()+'}]')[z](/\n/g,'},{')[z](/;/g,','));for(‌​i in p){l=p[i];c=0;for(k in l){if(!c){c=k;r[c]=0;}else{r[c]-=l[k];}};for(j in p){w=p[j][c];if(w!=null)r[c]+=w}};alert(JSON.stringify(r)[z](/"|{|}/g,'')[z](/,/‌​g,'\n'))? –  user2428118 Mar 28 at 13:35
    
Woah this saved me 7*4-(3*4+11) characters! (I also put (w!=null) as (!w) –  Antonio Ragagnin Mar 28 at 13:43
    
@AntonioRagagnin Could you please show your output? –  WallyWest Mar 31 at 3:48
    
Thanks for your message @WallyWest. Turns out !w was not a good idea to check w!=null and the script wasn't working anymore :p. Now I'll update it with the results –  Antonio Ragagnin Mar 31 at 6:29
    
Try using: z="replace";r={};p=eval(("[{"+prompt()+"}]")[z](/\n/g,"},{")[z](/;/g,","));for(‌​i in p){l=p[i];c=0;for(k in l)c?r[c]-=l[k]:(c=k,r[c]=0);for(j in p)w=p[j][c],null!=w&&(r[c]+=w)}alert(JSON.stringify(r)[z](/"|{|}/g,"")[z](/,/g,"‌​\n")) for 229 bytes... What I've done here is reduced the if(!c) sequence to a single ternary operator, and I've also incorporated it in its parent for loop... I've also done something similar with the other for loop... comma operators can work wonderfully to join multiple statements within a loop... –  WallyWest Mar 31 at 11:56

JavaScript(ES6) 175,166, 161, 156, 153147

Golfed

R={};prompt().split(/\s/).map(l=>{a=l.split(/[;,:]/);c=b=a[0];a.map(v=>b=!+v?v:(R[b]=(R[b]||0)+ +v c==b?b:R[c]-=+v))});for(x in R)alert(x+':'+R[x])

Ungolfed

R = {};
prompt().split(/\s/).map(l => {
    a = l.split(/[;,:]/);       // Split them all!! 
                                // Now in a we have big array with Country/Value items
    c = b = a[0];               // c - is first country, b - current country
    a.map(v =>                
         b = !+v ? v                 // If v is country (not a number), simply update b to it's value          
                 : (R[b] = (R[b] ||0) + +v   // Safely Add value to current country
                   c == b ? c : R[c] -= +v)  // If current country is not first one, remove debth 
    )
});
for (x in R) alert(x + ':' + R[x])

Output

US:9439.299999999997
FR:2598.8999999999996
ES:852.1
JP:4817.4
DE:2903.7
UK:1546.2000000000003
IT:887.5
PT:90.09999999999998
IE:48
GR:116.8
share|improve this answer
    
Not sure that ungolfed variant will be working correctly because in golfed variant i'm using one-line operators –  tt.Kilew Mar 31 at 9:27
    
can you please show your output? –  WallyWest Mar 31 at 12:04
1  
Replaced R[b] ? R[b] += +v : R[b] = +v to R[b]=R[b]||0+ +v –  tt.Kilew Mar 31 at 13:06
1  
Removed index i=0;i++%2==0?b=v to b=isNaN(+v)?v: –  tt.Kilew Mar 31 at 13:21
1  
isNaN(+v) -> !+v –  tt.Kilew Mar 31 at 13:28

Groovy 315

def f(i){t=[:];i.eachLine(){l=it.split(/;|,/);s=l[0].split(/:/);if(!z(s[0]))t.put(s[0],0);t.put(s[0],x(z(s[0]))+x(s[1]));(1..<l.size()).each(){n=l[it].split(/:/);t.put(s[0],x(z(s[0]))-x(n[1]));if(!z(n[0]))t.put(n[0],0);t.put(n[0],x(z(n[0]))+x(n[1]))}};t.each(){println it}};def x(j){j.toDouble()};def z(j){t.get(j)}

Output:
US=9439.299999999997
FR=2598.8999999999996
ES=852.1
JP=4817.4
DE=2903.7
UK=1546.2000000000003
IT=887.5
PT=90.09999999999998
IE=48.0
GR=116.8

Ungolfed:

input = """US:10800;FR:440.2,ES:170.5,JP:835.2,DE:414.5,UK:834.5
FR:1800;IT:37.6,JP:79.8,DE:123.5,UK:227,US:202.1
ES:700;PT:19.7,IT:22.3,JP:20,DE:131.7,UK:74.9,US:49.6,FR:112
PT:200;IT:2.9,DE:26.6,UK:18.9,US:3.9,FR:19.1,ES:65.7
IT:1200;JP:32.8,DE:120,UK:54.7,US:34.8,FR:309,ES:29.5
IE:200;JP:15.4,DE:82,UK:104.5,US:39.8,FR:23.8
GR:200;DE:15.9,UK:9.4,US:6.2,FR:41.4,PT:7.5,IT:2.8
JP:4100;DE:42.5,UK:101.8,US:244.8,FR:107.7
DE:2400;UK:141.1,US:174.4,FR:205.8,IT:202.7,JP:108.3
UK:1700;US:578.6,FR:209.9,ES:316.6,IE:113.5,JP:122.7,DE:379.3"""

ungolfed(input)

def ungolfed(i){
    def tallyMap = [:]
    i.eachLine(){ 
        def lineList = it.split(/;|,/)
        def target = lineList[0].split(/:/)

        if(!tallyMap.get(target[0])){tallyMap.put(target[0],0)}
        tallyMap.put(target[0],tallyMap.get(target[0]).toDouble() + target[1].toDouble())
        (1..lineList.size()-1).each(){ e ->
            def nextTarget = lineList[e].split(/:/)
            //subtract the debt
            tallyMap.put(target[0], (tallyMap.get(target[0]).toDouble() - nextTarget[1].toDouble()))
            //add the debt
            if(!tallyMap.get(nextTarget[0])){ tallyMap.put(nextTarget[0], 0) }
            tallyMap.put(nextTarget[0], (tallyMap.get(nextTarget[0]).toDouble() + nextTarget[1].toDouble()))  
        }
    }
    tallyMap.each(){
        println it
    }
}
share|improve this answer
    
Do you happen to have a link to where I can find more info on Groovy? –  WallyWest Mar 31 at 3:46
    
@WallyWest: I have this book, and learned tons from it. I feel that this is one of those languages that it is good to have a reference on the shelf. link, Also tons of info here: link –  md_rasler Apr 1 at 3:00

php - 333

$a='';while(($l=trim(fgets(STDIN)))!='')$a.=$l.'\n';$a=rtrim($a,'\n');$p=explode('\n',$a);foreach($p as $q){preg_match('/^([A-Z]+)/',$q,$b);preg_match_all('/'.$b[0].':(\d+(?:\.\d+)?)/',$a,$c);$e=ltrim(strstr($q,';'),';');preg_match_all('/([A-Z]+)\:(\d+(?:\.\d+)?)/',$e,$d);echo $b[0].':'.(array_sum($c[1])-array_sum($d[2])).PHP_EOL;}

Ungolfed version :

$a='';
while(($l=trim(fgets(STDIN)))!='')
    $a .= $l.'\n';
$a = rtrim($a,'\n');
$p = explode('\n',$a);
foreach($p as $q){
    preg_match('/^([A-Z]+)/',$q,$b);
    preg_match_all('/'.$b[0].':(\d+(?:\.\d+)?)/',$a,$c);
    $e = ltrim(strstr($q,';'),';');
    preg_match_all('/([A-Z]+)\:(\d+(?:\.\d+)?)/', $e, $d);
    echo $b[0].':'.(array_sum($c[1])-array_sum($d[2])).PHP_EOL;
}
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