maximum et minimum sont à la fois mal aussi Maxloc et Minloc sont à la fois mal l'impression de repos correctement seulement ces deux pièces

Question

program Loans(Input,Output); 
    uses crt; 
Var 
    i,m,l:Integer; 
    Month:Integer; 
    LoanID:String; 
    QUARTERS:array[1..4] of Integer; 
    max:integer; 
    min:integer; 
    maxloc:integer; 
    minloc:integer; 

begin 
    clrscr; 
    For i:=1 to 4 do 
    QUARTERS[m]:=0; 

    For l:=1 to 8 do 
begin 
    Writeln('Enter LoanID and Month'); 
    Readln(LoanID); 
    Readln(Month); 

    if Month <=3 then 
    QUARTERS[1]:=QUARTERS[1]+1 
    else if Month <=6 then QUARTERS[2]:=QUARTERS[2]+1 
    else if Month <=9 then QUARTERS[3]:=QUARTERS[3]+1 
    else QUARTERS[4]:=QUARTERS[4]+1 
    end; 

    max:= 0; 
    min:= 999; 
    for m:= 1 to 4 do 
    begin 
    if QUARTERS[m]>max then 
    max:=QUARTERS[m]; 
    maxloc:=m; 
    end; 
    begin 
    if QUARTERS[m]<min then 
    min:=QUARTERS[m]; 
    minloc:=m; 
    end; 

    for m:=1 to 4 do 
    Writeln('Quarter ',m,' has ',QUARTERS[m],' loans'); 

    Writeln('The maximum number of loans taken is ', max); 
    Writeln('The maximum number of loans were taken in quarter ', maxloc); 

    Writeln('The minimum number of loans taken is ', min); 
    Writeln('The minimum number of loans were taken in quarter ', minloc);`enter code here` 

end. 

Answer 1

Vous devez certainement formater le code correctement - alors vous verrez vos erreurs avec begin/end

for m:= 1 to 4 do 
begin 
    if QUARTERS[m]>max then begin 
    max:=QUARTERS[m]; 
    maxloc:=m; //update maxloc here!!! 
    end; //if 
    if QUARTERS[m]<min then 
    begin 
    min:=QUARTERS[m]; 
    minloc:=m; 
    end;//if 
end;//for m 

et vous pouvez changer tout votre code de quartiers

if Month <=3 then 
    QUARTERS[1]:=QUARTERS[1]+1 
else if Month <=6 then QUARTERS[2]:=QUARTERS[2]+1 
else if Month <=9 then QUARTERS[3]:=QUARTERS[3]+1 
else QUARTERS[4]:=QUARTERS[4]+1 
end; 

à une seule ligne:

Inc(QUARTERS[1 + (Month - 1) div 4]);