Tantangan

Diberikan dua input, ion positif dan ion negatif, Anda harus mengeluarkan rumus untuk senyawa ionik yang akan dibuat dari dua ion. Ini pada dasarnya berarti menyeimbangkan biaya sehingga mereka sama dengan nol.

Jangan repot-repot memformat rumus dengan angka subskrip, tetapi Anda harus memiliki tanda kurung untuk ion multi-atom (seperti NO3).

Anda tidak harus memperhitungkan kesalahan apa pun (misalnya, jika seseorang memasukkan dua ion negatif, Anda dapat membiarkan programnya gagal).

Catatan: Ambil Fe untuk memiliki muatan 3+

Ion

Semua ion yang perlu dipertanggungjawabkan ditemukan bersama dengan muatannya pada bagian kedua dari Lembar Data Kimia AQA GCSE .

Ion positif

• H ⁺

• Na ⁺

• Ag ⁺

• K ⁺

• Li ⁺

• NH ₄ ⁺

• Ba ²⁺

• Ca ²⁺

• Cu ²⁺

• Mg ²⁺

• Zn ²⁺

• Pb ²⁺

• Fe ³⁺

• Al ³⁺

Ion negatif

• Cl ^-

• Br ^-

• F ^-

• Saya ^-

• OH ^-

• TIDAK ₃ ^-

• O ^2-

• S ^2-

• SO ₄ ^2-

• CO ₃ ^2-

Contohnya

Beberapa contoh:

Pengembalian H dan O : -H2O

Pengembalian Ca dan CO ₃ : -CaCO3

Pengembalian Al dan SO ₄ : -Al2(SO4)3

Perhatikan kasus berikut yang harus Anda pertanggungjawabkan:

Pengembalian H dan OH : - H2O tidak H(OH)

CJam, 176 byte

"O S SO4 CO3 ""Cl Br F I OH NO3 """"H Na Ag K Li NH4 ""Ba Ca Cu Mg Zn Pb ""Fe Al "]_2{r\1$S+f#Wf=0#((z@}*_2$*_4=2*-_@/\@/@\]2/{~_({1$1>_el={'(@@')\}|0}&;}/]s"HOHH"1$#){;"H2O"}&

Cobalah online

Ini agak menyakitkan, terutama dengan semua kasus khusus untuk kurung, menunjukkan jumlah, H2O, dll.

Data tidak dalam format yang sangat ringkas. Bisa dipangkas lebih banyak, tetapi kode yang diperlukan untuk menafsirkannya mungkin akan mengkompensasi penghematan. Jadi saya pergi dengan array string, di mana setiap string berisi atom dengan muatan yang sama, dipesan dari -2 hingga +3 (di mana string untuk 0 kosong).

Penjelasan:

[..]  Data, as explained above.
_     Duplicate data, will need it for both inputs.
2{    Loop over two inputs.
  r     Get input.
  \     Swap data to top.
  1$    Copy input to top (keep original for later).
  S+    Add a space to avoid ambiguity when searching in data.
  f#    Search for name in all strings of data.
  Wf=   Convert search results to truth values by comparing them to -1.
  0#    Find the 0 entry, which gives the index of the matching string.
  ((    Subtract 2, to get charge in range [-2, 3] from index.
  z     Absolute value, we don't really care about sign of charge.
  @     Swap second copy of data table to proper position for next input.
}*    End loop over two inputs.
_2$*  Multiply the two charges.
_4=   We need the LCM. But for the values here, only product 4 is not the LCM.
2*-   So change it to 2.
_@/   Divide LCM by first charge to get first count.
\@/   Divide LCM by second charge to get second count.
@\]2/ Make pairs of name/count for both ions...
{     ... and loop over the pairs.
  ~     Unpack the pair.
  _(    Check if count is > 1.
  {       Handle count > 1.
    1$    Get copy of name to top of stack.
    1>    Slice off first character to check if rest contains upper case.
    _el   Create lower case copy.
    =     If they are different, there are upper case letters.
    {       Handle case where parentheses are needed.
      '(    Opening parentheses.
      @@    Some stack shuffling to get values in place.
      ')    Closing parentheses.
      \     And one more swap to place count after parentheses.
    }|    End parentheses handling.
    0     Push dummy value to match stack layout of other branch.
  }&    End count handling.
  ;     Pop unused count off stack.
}/    End loop over name/count pairs.
]s    Pack stack content into single string, for H2O handling.
"HOHH"  String that contains both HOH and OHH, which need to be H2O.
1$#)  Check if output is in that string.
{     If yes, replace with H2O.
  ;     Drop old value.
  "H2O" And make it H2O instead.
}&    End of H2O handling.

Lua , 174 242 byte

Saya lupa tanda kurung '-_-, yang membuat saya sampai 242. Oh well, setidaknya itu tantangan yang cukup menyenangkan.

i,c,a={Ba=2,Ca=2,Cu=2,Mg=2,Zn=2,Pb=2,Fe=3,Al=3,O=2,S=2,SO4=2,CO3=2},io.read(),io.read()
p,d=i[c]~=i[a],{SO4=1,NO3=1,OH=1,CO3=1}
k,m=p and i[c]or'',p and i[a]or''
a=k==m and a or (d[a]and'('..a..')'or a)
print(c..a=='HOH'and'H2O'or c..m..a..k)

Cobalah online!

Versi lama:

i,c,a={Ba=2,Ca=2,Cu=2,Mg=2,Zn=2,Pb=2,Fe=3,Al=3,O=2,S=2,SO4=2,CO3=2},io.read(),io.read()
p=i[c]~=i[a]
k,m=p and i[c]or'',p and i[a]or''
print(c..a=='HOH'and'H2O'or c..m..a..k)

Menyalahgunakan kecenderungan Lua untuk menginisialisasi segala sesuatu dengan nilai nol, kita dapat menghemat biaya penyimpanan. Meski begitu, Lua masih agak kikuk :(

Java ( 619 647 667 bytes)

[Memperbaiki] Pembaruan: H + OH mengembalikan HOH meskipun saya sulit mengkodekannya untuk .... mengerjakannya

Pembaruan [Tetap]: Terkadang Parenthesis muncul padahal seharusnya tidak

Kode

String f(String[]a){if(Arrays.equals(a,new String[]{"H","OH"})|Arrays.equals(a,new String[]{"OH","H"}))return "H2O";List<String>b=Arrays.asList(new String[]{"H","Na","Ag","K","Li","NH4","Ba","Ca","Cu","Mg","Zn","Pb","Fe","Al","Cl","Br","F","I","OH","NO3","O","S","SO4","CO3"});Integer[]c={1,1,1,1,1,1,2,2,2,2,2,2,3,3,1,1,1,1,1,1,2,2,2,2},d={5,18,19,22,23};List<Integer>j=Arrays.asList(d);int e=b.indexOf(a[0]),f=b.indexOf(a[1]),g=c[e],h=c[f],i;if(f<e){String p=a[0];a[0]=a[1];a[1]=p;i=g;g=h;h=i;i=e;e=f;f=i;}boolean k=j.contains(e),l=j.contains(f),m=g==h,n=g==1,o=h==1;return (k&!m&!o?"("+a[0]+")":a[0])+(m?"":h==1?"":h)+(l&!m&!n?"("+a[1]+")":a[1])+(m?"":g==1?"":g);}

Saya tidak yakin bagaimana melakukan ini tanpa pengkodean keras setiap ion, jadi itu lama. Beruntung karena semua muatannya adalah 1, 2, atau 3 sehingga menemukan jumlah masing-masing ion itu mudah.

Diperluas

import java.util.Arrays;
import java.util.List;
public class Compound {
    public static void main(String[]a){
        //System.out.println(f(a));
        String[] pos = new String[]{"H","Na","Ag","K","Li","NH4","Ba","Ca","Cu","Mg","Zn","Pb","Fe","Al"};
        String[] neg = new String[]{"Cl","Br","F","I","OH","NO3","O","S","SO4","CO3"};
        for(int i = 0; i < pos.length; i++){
            for(int j = 0; j < neg.length; j++){
                System.out.println(pos[i] + " + " + neg[j] + " = " + f(new String[]{pos[i],neg[j]}));
                System.out.println(neg[j] + " + " + pos[i] + " = " + f(new String[]{neg[j],pos[i]}));
            }
        }
    }
    static String f(String[]a){
        if(Arrays.equals(a,new String[]{"H","OH"})|Arrays.equals(a,new String[]{"OH","H"}))
            return "H2O";
        List<String>b=Arrays.asList(new String[]{"H","Na","Ag","K","Li","NH4","Ba","Ca","Cu","Mg","Zn","Pb","Fe","Al","Cl","Br","F","I","OH","NO3","O","S","SO4","CO3"});
        Integer[]c={1,1,1,1,1,1,2,2,2,2,2,2,3,3,1,1,1,1,1,1,2,2,2,2},d={5,18,19,22,23};
        List<Integer>j=Arrays.asList(d);
        int e=b.indexOf(a[0]),f=b.indexOf(a[1]),g=c[e],h=c[f],i;
        if(f<e){String p=a[0];a[0]=a[1];a[1]=p;i=g;g=h;h=i;i=e;e=f;f=i;}
        boolean k=j.contains(e),l=j.contains(f),m=g==h,n=g==1,o=h==1;
        return (k&!m&!o?"("+a[0]+")":a[0])+(m?"":o?"":h)+(l&!m&!n?"("+a[1]+")":a[1])+(m?"":n?"":g);
    }
}

Coba di sini

Data

Beri tahu saya jika ada yang salah

H + Cl = HCl
Cl + H = HCl
H + Br = HBr
Br + H = HBr
H + F = HF
F + H = HF
H + I = HI
I + H = HI
H + OH = H2O
OH + H = H2O
H + NO3 = HNO3
NO3 + H = HNO3
H + O = H2O
O + H = H2O
H + S = H2S
S + H = H2S
H + SO4 = H2SO4
SO4 + H = H2SO4
H + CO3 = H2CO3
CO3 + H = H2CO3
Na + Cl = NaCl
Cl + Na = NaCl
Na + Br = NaBr
Br + Na = NaBr
Na + F = NaF
F + Na = NaF
Na + I = NaI
I + Na = NaI
Na + OH = NaOH
OH + Na = NaOH
Na + NO3 = NaNO3
NO3 + Na = NaNO3
Na + O = Na2O
O + Na = Na2O
Na + S = Na2S
S + Na = Na2S
Na + SO4 = Na2SO4
SO4 + Na = Na2SO4
Na + CO3 = Na2CO3
CO3 + Na = Na2CO3
Ag + Cl = AgCl
Cl + Ag = AgCl
Ag + Br = AgBr
Br + Ag = AgBr
Ag + F = AgF
F + Ag = AgF
Ag + I = AgI
I + Ag = AgI
Ag + OH = AgOH
OH + Ag = AgOH
Ag + NO3 = AgNO3
NO3 + Ag = AgNO3
Ag + O = Ag2O
O + Ag = Ag2O
Ag + S = Ag2S
S + Ag = Ag2S
Ag + SO4 = Ag2SO4
SO4 + Ag = Ag2SO4
Ag + CO3 = Ag2CO3
CO3 + Ag = Ag2CO3
K + Cl = KCl
Cl + K = KCl
K + Br = KBr
Br + K = KBr
K + F = KF
F + K = KF
K + I = KI
I + K = KI
K + OH = KOH
OH + K = KOH
K + NO3 = KNO3
NO3 + K = KNO3
K + O = K2O
O + K = K2O
K + S = K2S
S + K = K2S
K + SO4 = K2SO4
SO4 + K = K2SO4
K + CO3 = K2CO3
CO3 + K = K2CO3
Li + Cl = LiCl
Cl + Li = LiCl
Li + Br = LiBr
Br + Li = LiBr
Li + F = LiF
F + Li = LiF
Li + I = LiI
I + Li = LiI
Li + OH = LiOH
OH + Li = LiOH
Li + NO3 = LiNO3
NO3 + Li = LiNO3
Li + O = Li2O
O + Li = Li2O
Li + S = Li2S
S + Li = Li2S
Li + SO4 = Li2SO4
SO4 + Li = Li2SO4
Li + CO3 = Li2CO3
CO3 + Li = Li2CO3
NH4 + Cl = NH4Cl
Cl + NH4 = NH4Cl
NH4 + Br = NH4Br
Br + NH4 = NH4Br
NH4 + F = NH4F
F + NH4 = NH4F
NH4 + I = NH4I
I + NH4 = NH4I
NH4 + OH = NH4OH
OH + NH4 = NH4OH
NH4 + NO3 = NH4NO3
NO3 + NH4 = NH4NO3
NH4 + O = (NH4)2O
O + NH4 = (NH4)2O
NH4 + S = (NH4)2S
S + NH4 = (NH4)2S
NH4 + SO4 = (NH4)2SO4
SO4 + NH4 = (NH4)2SO4
NH4 + CO3 = (NH4)2CO3
CO3 + NH4 = (NH4)2CO3
Ba + Cl = BaCl2
Cl + Ba = BaCl2
Ba + Br = BaBr2
Br + Ba = BaBr2
Ba + F = BaF2
F + Ba = BaF2
Ba + I = BaI2
I + Ba = BaI2
Ba + OH = Ba(OH)2
OH + Ba = Ba(OH)2
Ba + NO3 = Ba(NO3)2
NO3 + Ba = Ba(NO3)2
Ba + O = BaO
O + Ba = BaO
Ba + S = BaS
S + Ba = BaS
Ba + SO4 = BaSO4
SO4 + Ba = BaSO4
Ba + CO3 = BaCO3
CO3 + Ba = BaCO3
Ca + Cl = CaCl2
Cl + Ca = CaCl2
Ca + Br = CaBr2
Br + Ca = CaBr2
Ca + F = CaF2
F + Ca = CaF2
Ca + I = CaI2
I + Ca = CaI2
Ca + OH = Ca(OH)2
OH + Ca = Ca(OH)2
Ca + NO3 = Ca(NO3)2
NO3 + Ca = Ca(NO3)2
Ca + O = CaO
O + Ca = CaO
Ca + S = CaS
S + Ca = CaS
Ca + SO4 = CaSO4
SO4 + Ca = CaSO4
Ca + CO3 = CaCO3
CO3 + Ca = CaCO3
Cu + Cl = CuCl2
Cl + Cu = CuCl2
Cu + Br = CuBr2
Br + Cu = CuBr2
Cu + F = CuF2
F + Cu = CuF2
Cu + I = CuI2
I + Cu = CuI2
Cu + OH = Cu(OH)2
OH + Cu = Cu(OH)2
Cu + NO3 = Cu(NO3)2
NO3 + Cu = Cu(NO3)2
Cu + O = CuO
O + Cu = CuO
Cu + S = CuS
S + Cu = CuS
Cu + SO4 = CuSO4
SO4 + Cu = CuSO4
Cu + CO3 = CuCO3
CO3 + Cu = CuCO3
Mg + Cl = MgCl2
Cl + Mg = MgCl2
Mg + Br = MgBr2
Br + Mg = MgBr2
Mg + F = MgF2
F + Mg = MgF2
Mg + I = MgI2
I + Mg = MgI2
Mg + OH = Mg(OH)2
OH + Mg = Mg(OH)2
Mg + NO3 = Mg(NO3)2
NO3 + Mg = Mg(NO3)2
Mg + O = MgO
O + Mg = MgO
Mg + S = MgS
S + Mg = MgS
Mg + SO4 = MgSO4
SO4 + Mg = MgSO4
Mg + CO3 = MgCO3
CO3 + Mg = MgCO3
Zn + Cl = ZnCl2
Cl + Zn = ZnCl2
Zn + Br = ZnBr2
Br + Zn = ZnBr2
Zn + F = ZnF2
F + Zn = ZnF2
Zn + I = ZnI2
I + Zn = ZnI2
Zn + OH = Zn(OH)2
OH + Zn = Zn(OH)2
Zn + NO3 = Zn(NO3)2
NO3 + Zn = Zn(NO3)2
Zn + O = ZnO
O + Zn = ZnO
Zn + S = ZnS
S + Zn = ZnS
Zn + SO4 = ZnSO4
SO4 + Zn = ZnSO4
Zn + CO3 = ZnCO3
CO3 + Zn = ZnCO3
Pb + Cl = PbCl2
Cl + Pb = PbCl2
Pb + Br = PbBr2
Br + Pb = PbBr2
Pb + F = PbF2
F + Pb = PbF2
Pb + I = PbI2
I + Pb = PbI2
Pb + OH = Pb(OH)2
OH + Pb = Pb(OH)2
Pb + NO3 = Pb(NO3)2
NO3 + Pb = Pb(NO3)2
Pb + O = PbO
O + Pb = PbO
Pb + S = PbS
S + Pb = PbS
Pb + SO4 = PbSO4
SO4 + Pb = PbSO4
Pb + CO3 = PbCO3
CO3 + Pb = PbCO3
Fe + Cl = FeCl3
Cl + Fe = FeCl3
Fe + Br = FeBr3
Br + Fe = FeBr3
Fe + F = FeF3
F + Fe = FeF3
Fe + I = FeI3
I + Fe = FeI3
Fe + OH = Fe(OH)3
OH + Fe = Fe(OH)3
Fe + NO3 = Fe(NO3)3
NO3 + Fe = Fe(NO3)3
Fe + O = Fe2O3
O + Fe = Fe2O3
Fe + S = Fe2S3
S + Fe = Fe2S3
Fe + SO4 = Fe2(SO4)3
SO4 + Fe = Fe2(SO4)3
Fe + CO3 = Fe2(CO3)3
CO3 + Fe = Fe2(CO3)3
Al + Cl = AlCl3
Cl + Al = AlCl3
Al + Br = AlBr3
Br + Al = AlBr3
Al + F = AlF3
F + Al = AlF3
Al + I = AlI3
I + Al = AlI3
Al + OH = Al(OH)3
OH + Al = Al(OH)3
Al + NO3 = Al(NO3)3
NO3 + Al = Al(NO3)3
Al + O = Al2O3
O + Al = Al2O3
Al + S = Al2S3
S + Al = Al2S3
Al + SO4 = Al2(SO4)3
SO4 + Al = Al2(SO4)3
Al + CO3 = Al2(CO3)3
CO3 + Al = Al2(CO3)3

Catatan

Saya mulai di Pyth, tapi kemudian saya merasa terganggu dengan urutan dan tanda kurung, inilah yang saya miliki jika ada yang ingin menyelesaikannya.

=G["H" "Na" "Ag" "K" "Li" "NH4" "Ba" "Ca" "Cu" "Mg" "Zn" "Pb" "Fe" "Al" "Cl" "Br" "F" "I" "OH" "NO3" "O" "S" "SO4" "CO3" 1 1 1 1 1 1 2 2 2 2 2 2 3 3 1 1 1 1 1 1 2 2 2 2)J@G+24xG@QZK@G+24xG@Q1@QZ?kqJKK@Q1?kqJKJ

JavaScript (ES6), 316 277 byte

Saya telah mengglobal pdan nvariabel (seperti yang saya lakukan di CoffeeScript) untuk pengujian lebih mudah. Melokalkan variabel tidak akan membuat perbedaan dalam jumlah karakter.

f=(x,y)=>{i='indexOf',d='~NH4KNaAgLiBa~CaCuMgZnPbFeAlSO4CO3ClBrFIOHNO3',k=d[i](x),l=d[i](y),a=k<28?x:y,b=k<28?y:x,r=Math.ceil(k/11)-1,s=b=='F'?0:l<32;if(r==s)r=s=0;a=s&&a=='NH4'?'(NH4)':a;b=r&&/[A-Z]{2}/.test(b)?`(${b})`:b;return'H'==a&&b=='OH'?'H2O':a+(s?s+1:'')+b+(r?r+1:'')}


// Original attempt, 316 bytes
p={H:1,Na:1,Ag:1,K:1,Li:1,NH4:1,Ba:2,Ca:2,Cu:2,Mg:2,Zn:2,Pb:2,Fe:3,Al:3},n={Cl:1,Br:1,F:1,I:1,OH:1,NO3:1,O:2,S:2,SO4:2,CO3:2},f=(x,y)=>{a=p[x]?x:y,b=p[x]?y:x,z=p[a]==2&&n[b]==2,r=+z||n[b],s=+z||p[a];a=--r&&a=='NH4'?'(NH4)':a;b=--s&&/[A-Z]{2}/.test(b)?`(${b})`:b;return'H'==a&&b=='OH'?'H2O':a+(r?r+1:'')+b+(s?s+1:'')}

Varian ES5, 323 284 byte

Tidak banyak perubahan selain menghilangkan fungsi panah dan string templat:

f=function(x,y){i='indexOf',d='~NH4KNaAgLiBa~CaCuMgZnPbFeAlSO4CO3ClBrFIOHNO3',k=d[i](x),l=d[i](y),a=k<28?x:y,b=k<28?y:x,r=Math.ceil(k/11)-1,s=b=='F'?0:l<32;if(r==s)r=s=0;a=s&&a=='NH4'?'(NH4)':a;b=r&&/[A-Z]{2}/.test(b)?'('+b+')':b;return'H'==a&&b=='OH'?'H2O':a+(s?s+1:'')+b+(r?r+1:'')}


// Original attempt, 323 bytes
p={H:1,Na:1,Ag:1,K:1,Li:1,NH4:1,Ba:2,Ca:2,Cu:2,Mg:2,Zn:2,Pb:2,Fe:3,Al:3},n={Cl:1,Br:1,F:1,I:1,OH:1,NO3:1,O:2,S:2,SO4:2,CO3:2},f=function(x,y){a=p[x]?x:y,b=p[x]?y:x,z=p[a]==2&&n[b]==2,r=+z||n[b],s=+z||p[a];a=--r&&a=='NH4'?'(NH4)':a;b=--s&&/[A-Z]{2}/.test(b)?'('+b+')':b;return'H'==a&&b=='OH'?'H2O':a+(r?r+1:'')+b+(s?s+1:'')}

f=function(x,y){i='indexOf',d='~NH4KNaAgLiBa~CaCuMgZnPbFeAlSO4CO3ClBrFIOHNO3',k=d[i](x),l=d[i](y),a=k<28?x:y,b=k<28?y:x,r=Math.ceil(k/11)-1,s=b=='F'?0:l<32;if(r==s)r=s=0;a=s&&a=='NH4'?'(NH4)':a;b=r&&/[A-Z]{2}/.test(b)?'('+b+')':b;return'H'==a&&b =='OH'?'H2O':a+(s?s+1:'')+b+(r?r+1:'')}

;['H', 'Na', 'Ag', 'K', 'Li', 'NH4', 'Ba', 'Ca', 'Cu', 'Mg', 'Zn', 'Pb', 'Fe', 'Al'].forEach(function (q) {
    ['Cl', 'Br', 'F', 'I', 'OH', 'NO3', 'O', 'S', 'SO4', 'CO3'].forEach(function (w) {
        document.body.innerHTML += '<p>' + q + ' + ' + w + ' = ' + f(q, w);
    });
});

CoffeeScript, 371 333 byte

Hitungan ini termasuk baris baru (beberapa baris baru dapat diganti dengan titik koma tetapi itu tidak akan mempengaruhi jumlah karakter)

f=(x,y)->(i='indexOf';d='~NH4KNaAgLiBa~CaCuMgZnPbFeAlSO4CO3ClBrFIOHNO3';k=d[i] x;r=-1+Math.ceil k/11;l=d[i] y;a=if k<28then x else y
b=if k<28then y else x
s=if b=='F'then 0else 32>l
r=s=0if r==s;a='(NH4)'if'NH4'==a&&s;b='('+b+')'if/[A-Z]{2}/.test(b)&&r;if'H'==a&&b=='OH'then'H2O'else a+(if!s then''else 1+s)+b+(if!r then''else 1+r))

# Original attempt, 371 bytes
p={H:1,Na:1,Ag:1,K:1,Li:1,NH4:1,Ba:2,Ca:2,Cu:2,Mg:2,Zn:2,Pb:2,Fe:3,Al:3}
n={Cl:1,Br:1,F:1,I:1,OH:1,NO3:1,O:2,S:2,SO4:2,CO3:2}
f=(x,y)->(a=if p[x]then x else y
b=if p[x]then y else x
z=p[a]==n[b]==2;r=+z||n[b];s=+z||p[a];if--r&&a=='NH4'then a='(NH4)'
if--s&&/[A-Z]{2}/.test(b)then b='('+b+')'
if'H'==a&&b=='OH'then'H2O'else a+(if!r then''else r+1)+b+(if!s then''else s+1))

CJam (137 bytes)

{{"SO4CO3ClBrFINO3OHNaAgKLiNH4BaCaCuMgZnPbFeAl":I\#~}$_s"HOH"="HO"1/@?_{I\#G-_0<B*-B/_W>+z}%_~1$=\1?f/W%[.{:T1>{_{'a<},,1>{'(\')}*T}*}]s}

Ini adalah blok anonim (fungsi) yang mengambil dua ion sebagai string yang terbungkus dalam daftar, dan mengembalikan sebuah string.

Demo online .

Pembedahan

{                        e# Begin a block
  {                      e#   Sort the input list
    "SO4...Al"           e#     Ions in ascending order of charge
    :I                   e#     Stored as I for future reuse
    \#~                  e#     Index and bit-invert to sort descending
  }$

  _s"HOH"="HO"1/@?       e#   Special case for water: replace ["H" "OH"] with ["H" "O"]

  _{I\#G-_0<B*-B/_W>+z}% e#   Copy the list and hash index in I to find the charges
  _~1$=\1?f/             e#   Replace [2 2] by [1 1]
  W%                     e#   Reverse the charges
  [                      e#   Gather in an array
    .{                   e#   Pairwise for each ion and its opponent's reduced charge...
      :T1>{              e#     If the charge (copied to T) is greater than 1
        _{'a<},,         e#       Count the characters in the ion which are before 'a'
        1>{'(\')}*       e#       If there's more than one, add some parentheses
        T                e#       Append T
      }*
    }
  ]
  s                      e#   Flatten the list to a single string
}

Python 3 , 364 byte

Menyimpan ion dengan nilai absolut dari muatannya seperti index+1dalam larik 2-D (elemen 0-indeks memiliki muatan + - 1, dll.). Menggunakan string.split () untuk menyimpan beberapa karakter di sana. Menangani kasus khusus H + OH = H2Opertama, lalu menghitung berapa banyak ion yang dibutuhkan dari masing-masing jenis sebagai LCM dari dua muatan mereka dibagi dengan muatannya. Kemudian tambahkan tanda kurung jika diperlukan serta jumlah ion yang sebenarnya.

from math import*
o=["NH4 NO3 H Na Ag K Li OH I F Br Cl".split(),"CO3 SO4 Ba Ca Cu Mg Zn Pb S O".split(),["Fe","Al"]]
p=["H","OH"]
def c(i):
 for h,k in enumerate(o):
  if i in k:return-~h
def b(x,y):
 if x in p and y in p:return"H20"
 i,j=c(x),c(y);g=gcd(i,j);i//=g;j//=g;return((x,"(%s)"%x)[x[-1]in"34"]+str(j),x)[j==1]+((y,"(%s)"%y)[y[-1]in"34"]+str(i),y)[i==1]

Cobalah online!