/// Original Code by Yash Thatte /// Solution to Long Calculation by $Vengat, /// on 21/9/2017... /// Input Requirement - Operand 1, Operator /// and Operand 2, all in /// a same line. Spaces /// not needed /// I just added the Multiplication and /// Division Part /// Modifier - Kinshuk Vasisht #include #include #include #include using namespace std; void space(int n) /// Print Spaces { for(int i=0;i9) cout<<" "; else cout<<" "; cout<0){ n/=10; a++; } return a; } int pow(int base,int expo) /// pow() for Integers { int res =1; for(int i=0;in3) { int no = tmp - n3; if(first) str.resize(len(n1)-len(no)); if(str.length()!=0) { no*=10; no+=(static_cast(str[str.length()-1])-48); str.resize(str.length()-1); } return no; } else l1--; } } void addt(int n1,int n2) /// Addition, + { int n1L = len(n1); int n2L = len(n2); int ans = n1 + n2; int ansL = len(ans); int *num1 = new int [ansL]; int *num2 = new int [ansL]; int *res = new int [ansL]; int *carry = new int [ansL]; for(int i=0;i=0;i--) { if((num1[i]+num2[i]+carry[i])>9) carry[i-1]=1; } printCarry(carry,ansL); cout<n1) swap(n1,n2); int n1L = len(n1); int n2L = len(n2); int ans = n1-n2; int ansL = len(ans); int *num1 = new int [n1L]; int *num2 = new int [n1L]; int *res = new int [n1L]; int *borr = new int [n1L]; for(int i=0;i=0;i--) { if(borr[i]2) h = i-1; else h=i; cout<>a>>op>>b; switch(op){ case '+': addt(a,b); break; case '-': subt(a,b); break; case '/': divt(a,b); break; case '*': mult(a,b); break; default: break; } cout<

Long Calculation - Updated