Question
A. Run the program advance20.ccp, which displays five rows of asterisks. Close the Command Prompt window. B. Modify the program to allow the user to
A. Run the program advance20.ccp, which displays five rows of asterisks. Close the Command Prompt window.
B. Modify the program to allow the user to specify the outer loops ending and increment values. The ending value determines the maximum number of asterisks to display. The increment value determines the number of asterisks to repeat.
C. Run the program you made from B. Test the program by entering the numbers 4 and 1 as the maximum number of asterisks and the number of asterisks to repeat, respectively. The program should display four rows of asterisks as follows: one asterisk, two asterisks, three asterisks, and four asterisks.
D. Run the program again. This time, enter the numbers 9 and 3 as the maximum number of asterisks and the number of asterisks to repeat, respectively. The program should display three rows of asterisks as follows: three asterisks, six asterisks, and nine asterisks.
E. Run the program again. Enter 7 and 3 as the maximum number of asterisks and the number of asterisks to repeat, respectively. The program displays only two rows of asterisks. The first row contains the expected three asterisks, but the second row contains six asterisks rather than seven asterisks. This is because the maximum number of asterisks (7) is not evenly divisible by the number of asterisks to repeat (3). Modify the program so that it displays the asterisks only when the maximum number is evenly divisible by the number to repeat; otherwise, display the message The maximum number must be evenly divisible by the number to repeat.
F. Then run the program. Test the program three times, using the data from Steps c, d, and e. Make sure to take a snap-shot of your results and upload you code and results using the uplink.
//Advanced20.cpp - displays a pattern of asterisks //Created/revised byon #include using namespace std; int main() { for (int outer = 2; outer < 11; outer += 2) { for (int nested = 1; nested <= outer; nested += 1) cout << '*'; //end for cout << endl; } //end for return 0; } //end of main function
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