Question

Hello, I am having trouble with this C++ programming assignment. If someone could write the code or at least part of it that would help me, because every code I try has errors.

Using if Statements, Loops and Nested Loops • Scanning Characters in a String Using a Loop • Performing Character Arithmetic In project, you will write an interactive program that, counts the number of digits in a non-negative integer, factorizes the integer into powers of ten and its digits, computes the sum of its digits and counts the number of odd digits that it has. The output of the program will be formatted as shown in the sample runs. Reuse variables where possible and avoid declaring variables for values that are calculated and used only once.

Programming Tips and Suggestions

1. Use the standard C++ math library log10 function to determine the number of digits, d, in a positive integer, n. The number of digits in n is given by the expression d=static cast< int >(log10(n))+1. The place value of the most significant digit is 10d−1.

2. Use a loop to peel off the digits of an integer beginning with the most significant digit. Each non-zero digit should be expressed as a power of 10 as shown in the sample run. Observe that the most significant digit of a non-negative integer n is n/p, where p is the place value of the most significant digit. To obtain the second most significant digit, divide the number formed by peeling off the most significant digit from the number by the place value of its most significant digit. Use the pow only to obtain the most significant digit of the original number.

3. The digits should be summed and the number of odd digits in the number should be counted. Use the modulus operator when counting odd digits rather than an if statement that checks whether the digits 1, 3, 5, 7 or 9.

4. An if statement will be required to avoid printing a power of 10 when the digit is a 0.

5. Only one loop should be used for this program.

6. When factorizing the integer into powers of 10 using its digits, you may want to deal with three separate cases: 1-digit, 2-digit and 3-or-moredigit cases. For the 3-or-more-digit case, process the first digit before the loop, process all but the last two digits within the loop and the last two digits after the loop.

7. Program incrementally. Get the program to work for one integer before adding a loop for it to work for any number of integers. First, get the factorization to work, then the summing of the digits and then the counting of the odd digits.

Here is a hand-trace of how to obtain the digits of a non-zero number. It consists of alternating the integer division and integer modulo operations while updating the place value, by dividing it by 10, as digits are extracted from the number from the most significant to the units digit. number = 962357 number of digits = 6 Most significant digit’s place value = 10^(6-1) = 10^5 = 100000 Digit Place Value Remainder 962357/100000 = 6 10^5 962357 (mod 100000) = 62357 62357/10000 = 6 10^4 62357 (mod 10000) = 2357 2357/1000 = 2 10^3 2357 (mod 1000) = 357 357/100 = 3 10^2 357 (mod 100) = 57 57/10 = 5 10 57 (mod 10) = 7 7 1

Here are a few sample runs.

Listing 1: A Sample Run 1 Enter an integer -> -2 The integer must be non-negative.

Listing 2: A Sample Run 2 Enter an integer -> 104529 104529 = 1 x 10^5 + 4 x 10^3 + 5 x 10^2 + 2 x 10 + 9 sumOfDigits (104529) = 21 #OddDigits (104529) = 3

Listing 3: A Sample Run 3 Enter an integer -> 1005 1005 = 1 x 10^3 + 5 sumOfDigits (1005) = 1 + 0 + 0 + 5 = 6 #OddDigits(1005) = 2

Listing 4: A Sample Run 4 Enter an integer -> 7654321 7654321 = 7 x 10^6 + 6 x 10^5 + 5 x 10^4 + 4 x 10^3 ← + 3 x 10^2 + 2 x 10 + 1 sumOfDigits (7654321) = 28 #OddDigits (7654321) = 4

Listing 5: A Sample Run 5 Enter an integer -> 129 129 = 1 x 10^2 + 2 x 10 + 9 sumOfDigits(129) = 12 #OddDigits(129) = 2

Listing 6: A Sample Run 6 Enter an integer -> 5 5 = 5 sumOfDigits(5) = 5 #OddDigits(5) = 1

Listing 7: A Sample Run 7 Enter an integer -> 20 20 = 2 x 10 sumOfDigits(20) = 2 #OddDigits(20) = 0 Listing 8: A Sample Run 8 Enter an integer -> 0 0 = 0 sumOfDigits(0) = 0 #OddDigits(0) = 0 Listing 9: A Sample Run 9 Enter an integer -> 10100 10100 = 1 x 10^4 + 1 x 10^2 sumOfDigits (10100) = 2 #OddDigits (10100) = 2

Answer 1

Thanks for the question, here is the complete code in C++. Hope it helps you. Let me know for any help with any other questions. Thanks a lot, do give a thumbs up please : )

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#include<iostream>
#include<cmath>

using namespace std;

int main(){
  
   int n;
   printf("Enter an integer: "); cin>>n;
   if(n<0){
      cout<<"The integer must ne non-negative.";
       return 0;
   }  
   int digits=static_cast<int>(log10(n))+1;
  
   int digitSum =0;
   int divisor=0;
   cout<<n<<" = ";
   int temp=n;
   int oddDigits=0;
   for(int digit=digits-1; digit>=0; digit--){
       divisor = static_cast<int>(pow(10,digit));
       int msd = n/divisor;
       n = n%divisor;
       cout<<msd<<"x"<<10<<"^"<<digit;
       if(digit!=0) cout<<" + ";
       digitSum+=msd;
       oddDigits += msd%2==1?1:0;
   }
   cout<<"\nSum of Digits ("<<temp<<") = "<<digitSum<<endl;
       cout<<"Odd Digits ("<<temp<<") = "<<oddDigits<<endl;
  
  
}

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