ProcedurProgramavimas_2.15

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
long double n,ats,d=1,ch=1,sk0,sk1,sk2;
long double aa,ab,ac,ad,ia,ib,ic,id,qa,qb,qc,qd;
long long int rez;
char mas[4],mas2[4];
cin>>mas[0]>>mas[1]>>mas[2]>>mas[3]>>n;
aa = mas[0] - '0';
ab = mas[1] - '0';
ac = mas[2] - '0';
ad = mas[3] - '0';
sk0=(aa*1000)+(ab*100)+(ac*10)+(ad);ats=sk0;
if(n==ch){cout<<ats<<endl;return 0;}

mn:
mas2[0]=mas[0];mas2[1]=mas[1];mas2[2]=mas[2];mas2[3]=mas[3];
//Nedidejancia tvarka (sk1)
for(long long int i = 4 - 1; i >= 1; i--){
   for(long long int j = 0; j < i; j++)
   {
        if(mas[j] > mas[j+1])
        {
             int foo = mas[j];
             mas[j] = mas[j+1];
             mas[j+1] = foo;
        }
   }
}
ia = mas[0] - '0';
ib = mas[1] - '0';
ic = mas[2] - '0';
id = mas[3] - '0';
sk1=(ia*1000)+(ib*100)+(ic*10)+(id);
//Nemazejancia tvarka (sk2)
for(long long int i = 4 - 1; i >= 1; i--){
   for(long long int j = 0; j < i; j++)
   {
        if(mas2[j] < mas2[j+1])
        {
             int foo = mas2[j];
             mas2[j] = mas2[j+1];
             mas2[j+1] = foo;
        }
   }
}
qa = mas2[0] - '0';
qb = mas2[1] - '0';
qc = mas2[2] - '0';
qd = mas2[3] - '0';
sk2=(qa*1000)+(qb*100)+(qc*10)+(qd);
//Isvedam atsakyma
rez=sk2-sk1;
mas[0]='0' + (rez/1000);
mas[1]='0' + ((rez-((rez/1000)*1000))/100);
mas[2]='0' + (((rez-((rez/1000)*1000))-(((rez-((rez/1000)*1000))/100)*100))/10);
mas[3]='0' + (((rez-(rez/1000)*1000)-(((rez-((rez/1000)*1000))/100)*100))-((((rez-((rez/1000)*1000))-(((rez-((rez/1000)*1000))/100)*100))/10)*10));
ats=rez;
ch+=1;
    if(n==ch || ch>10000){cout<<setprecision(60)<<ats<<endl;return 0;}else{goto mn;}
return 0;
}