Hello I'm trying to display only the amount of odd/even numbers for the digits entered. I've tried multiple methods but failed to find any solution. This is the problem and what I have so far.
Write a program that allows the user to enter 10 separate whole numbers. After accepting these 10 numbers from the user, the program should display output to the user informing them how many of the numbers entered were odd numbers and how many were even numbers.
#include <iostream>
using namespace std;
int main() {
char even, odd;
int number;
for(int i = 1;i<=10;i++) {
cout << "Enter Number " << i << ":" ;
i=i+0;
cin >> number ;
}
if (number%2==0){
number = even;
}
cout<< "You entered:\n";
cout << "Odd Numbers: " << odd << endl;
cout << "Even Numbers: " << even << endl;
return 0;
}
There are a few things in your code that do not look right.
for(int i = 1; i <= 10; i++)
{
cout << "Enter Number " << i << ":";
i = i + 0;
cin >> number;
}
What are you hoping to accomplish with the line i = i + 0? Your loop will work just fine without it.
char even, odd;
Technically, since char is a numeric type, you may keep track of the number of even and odd numbers encountered by keeping track of them. However, you aren't doing that.
The statement:
if (number%2==0){
number = even;
}
Is saying that if the input number is even, assign the current value of char even to number. However, this doesn't make sense, since you never stored a value in even earlier. Also, you're doing this outside the loop, so effectively only the last of the 10 values read into number would ever be counted in your calculations (if you had that done correctly).
What you should do:
int even = 0, odd = 0;
for(...)
{
// read input into "number"...
if(number % 2 == 0)
{
even++;
}
else
{
odd++;
}
}
Related
I am trying to get this 'do while' loop to run 3 times and then display the amount in the accumulator contain within a while loop inside the 'do while' loop.
It seems to be counting correctly, but only runs the while loop on the first. When run, instead of going on to ask for the next set of numbers, it just displays the first batch (added up correctly). I have tried switching some of the code around and searching google, but can't find the answer.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int storeNum = 1;
int payRollAmount = 0;
int totalPayroll = 0;
do
{
cout << "Store " << storeNum << ":" << endl;
while (payRollAmount <= -1)
{
cout << "Enter Store's Payroll Amount (-1 to exit): ";
cin >> payRollAmount;
totalPayroll += payRollAmount;
}
storeNum++;
} while (storeNum <= 3);
cout << "The Total Payroll is: " << totalPayroll << endl;
system("pause");
return 0;
}
The code should take in an unknown amount of "payrolls," allow you to exit using -1, and then continue on to the next stores payrolls. It should do this 3 times, and then display the total amount (all numbers entered added together.
Hi perhaps reset payRollAmount at each iteration? That way it will continue to request the input.
for (int amount = 0; amount != -1; ) {
cout << "Enter Store's Payroll Amount (-1 to exit): ";
cin >> amount;
totalPayroll += amount;
}
I am writing code for an assignment that wants me to make a program that asks the user for the amount of integers they'd like to input then it accepts each input while testing if the value is the max value or the minimum.
The program runs fine but for some reason will stop and show me the min and max number when I have entered in 1 integer less than the original input.
Examples of the problem I am having:
If I input 5 for the first value, it asks me to enter 5 integers.
After entering 4 numbers, 1 2 3 and 4.
It tells me the max is 4 and min is 1.
It prevents me from entering in the 5th integer.
Additionally,
If I input 5 for the first value, it asks me to enter 5 integers.
If I enter a negative number, like -1, the input allowed to me is further
shortened.
If I enter -1, 2, 3 then the output is min: 2 and max: 3
I have two main questions:
What adjustments do I need to make to my code so that I can properly
enter in the number of integers initially asked for?
What adjustments need to be made in order for me to be able to enter
negative integers so they are outputted correctly?
The code:
#include <iostream>
using namespace std;
int main()
{
int input;
int tmp;
int counter = 1;
int max_num = 0;
int min_num;
// prompt user for integer amount
cout << "How many integers would you like to enter? " << endl;
cin >> input;
cout << "Please enter " << input << " integers." << endl;
tmp = input;
// loop for requested amount with a test for each input
while (counter <= tmp) {
cin >> input;
// if smaller than previous number it is the minimum
if (input < min_num || min_num == -1) {
min_num = input;
counter++;
}
// if larger than previous number it becomes max number else
if (input > max_num) {
max_num = input;
counter++;
} else { // continue loop if number isn't bigger than max or smaller than min
counter++;
}
}
// display the max and min
cout << "min: " << min_num << endl;
cout << "max: " << max_num << endl;
return 0;
}
Create a program that takes as input a string of digits and outputs
the sum of the even and the sum of the odd digits. This program is in
C++
NOTE: It is recommended that you use a "for" loop to iterate over the
sting of digits. HINT: Use the modulus operator to determine even or
odd digits.
A clever method of converting a "char" to an "int" is as follows link:
char a = '4';
int ia = a - '0';
The above code takes advantage of the character's position in the
ASCII table to convert it to an integer.
Below is the code I have so far:
int main() {
int sumOdd = 0; // For accumulating odd numbers, init to 0
int sumEven = 0; // For accumulating even numbers, init to 0
int digits; // Sum from 1 to this upperbound
// Prompt user for an upperbound
cout << "Please enter a string comprised ONLY of digits:" << endl;
cout << endl;
cin >> digits;
// Use a while-loop to repeatedly add 1, 2, 3,..., to the upperbound
int number = 1;
while (number <= digits) {
if (number % 2 == 0) { // Even number
sumEven += number; // Add number into sumEven
} else { // Odd number
sumOdd += number; // Add number into sumOdd
}
++number; // increment number by 1
}
// Print the results
cout << "The string of digits \"" << digits << "\""
<< " contained "
<< "characters." << endl;
cout << "The sum of the even digits is: " << sumEven << endl;
cout << "The sum of the odd digits is: " << sumOdd << endl;
return 0;
}
This is my output compared to the output I need
Input:
1234567890
Output:
Please enter a string comprised ONLY of digits:
The string of digits "1234567890" contained characters.
The sum of the even digits is: -380436870
The sum of the odd digits is: -997720815
Expected output
Please enter a string comprised ONLY of digits:
The string of digits "1234567890" contained 10 characters.
The sum of the even digits is: 20
The sum of the odd digits is: 25
Overall I'm having trouble counting the input and getting the correct formula for my even and odd numbers. Any help is appreciated thank you so much!
A simple method is to keep the number in text form:
std::string number_as_text;
cout << "Enter number: ";
cin >> number_as_text;
This allows you to check each digit for even or odd:
const size_t length = number_as_text.length();
for (size_t i = 0; i < length; ++i)
{
char digit_character = number_as_text[i];
if (isdigit(digit_character))
{
if (digit_character % 2 == 0)
{
// digit is even
}
else
{
// digit is odd
}
}
}
If you don't like isdigit(), you can replace with:
if ((digit_character >= '0') && (digit_character < '9'))
An important note is that there is a difference between the textual representation of digits and the internal, numeric, representation of digits.
Edit 1: % on char type
The char data type is an integer. The remainder operator, %, works on integer types. Thus you can use % on char types.
Note: this operation assumes that the character mapping of '0' is an even integer and the other digits are successive in values.
Read the inputs in a string field and not an integer as -
std::string digits;
Then, just run a valid loop to find out the sum of the even digits and the odd digits -
for(int i=0; i<digits.length(); i++){
int digit = digits[i]-'0';
if(digit % 2 == 0)
sumEven+=digit;
else
sumOdd+=digit;
}
Below is a running piece of code.
// Example program
#include <iostream>
#include <string>
using namespace std;
int main() {
int sumOdd = 0; // For accumulating odd numbers, init to 0
int sumEven = 0; // For accumulating even numbers, init to 0
std::string digits; // Sum from 1 to this upperbound
// Prompt user for an upperbound
cout << "Please enter a string comprised ONLY of digits:" << endl;
cout << endl;
cin >> digits;
// Use a while-loop to repeatedly add 1, 2, 3,..., to the upperbound
int number = 1;
for(int i=0; i<digits.length(); i++){
int digit = digits[i]-'0';
if(digit % 2 == 0)
sumEven+=digit;
else
sumOdd+=digit;
}
// Print the results
cout << "The string of digits \"" << digits << "\""
<< " contained "
<< "characters." << endl;
cout << "The sum of the even digits is: " << sumEven << endl;
cout << "The sum of the odd digits is: " << sumOdd << endl;
return 0;
}
You got it wrong you are supposed to do sum of 1+3+5+7+9 and sum of 2+4+6+8+0. What are you doing instead is sum of all even numbers smaller than 1234567890 and sum of odd numbers smaller than 1234567890.
Maybe consider instead of getting all numbers in string by cin >> digits read it from input by characters one by one.
Check this code it reads characters from input and spits them back.
#include<iostream>
using namespace std;
int main() {
cout << "Please enter a string comprised ONLY of digits:" << endl;
cout << endl;
char one;
while ( true ) {
cin >> one;
// here will come your part with deciding even / odd and counting instead of pritnting out.
cout << " " << one;
if ( cin.peek() == '\n' ) { break; }
}
return 0;
}
And now it's your turn to put it together.
Use std::getline to read all digits into a std::string:
std::cout << "Enter string of numbers: " << std::endl;
if (std::string numbers; std::getline(std::cin, numbers)) {
/* ... */
}
Note the use of conditional-if above from the upcoming C++1z standard.
Then, you could make use of std::pair to neatly accumulate the odds and evens like this:
std::pair odd_even{0, 0};
for (auto c : numbers) {
(c % 2 ? std::get<0>(odd_even) : std::get<1>(odd_even)) += c - '0';
}
Live example
This one logically should work
for(int i=0; i<digits.lenght; i++){
if(digits[i] % 2 == 0)
sumEven+=digits[i];
else
sumOdd+=digits[i];
}
i have the next code which asks the user to introduce a number larger than 100000000, and then it asks for a digit that the code must search in the number, finally the code shows how many times the digit appears on the number, it seems to be easy but i have a restriction:
the data type cannot be a string or a char, thats why i am using an int, but when i introduce a real big number like 100100010000100 the code just doesn´t work properly, how could i solve this, any ideas???if someone could help me out with this i would appreciate it a lot
#include <iostream>
using namespace std;
int searchDigit(long int num,int digit);
int main()
{
long int num;
int digit,x;
cout << "Give me the number: " << endl;
cin >> num;
cout << "Digit " << endl;
cin >> digito;
x = searchDigit(num,digit);
cout << "\nThe digit " << digit << " appears " << x << " times" << endl;
return 0;
}
int searchDigit(long int num,int digit)
{
int r,c,p = 0;
for(c = num;c != 0;c = c/10)
{
r = c % 10;
if(r == digit)
p++;
}
return p;
}
Notice that in searchDigit you have c = num in the for-loop. If long int is larger than int - which is true on many platforms - you will lose the high bits of num.
If you enabled more compiler warnings, this would probably be picked up. It's hard to be more specific, since you haven't provided information about the platform, compiler, or the flags used.
My assignment is to create a program in which the user inputs a starting value and ending value. The program should then sum all the numbers within that range. In addition it should sum the odd numbers and the even numbers. My issue is determining which numbers in the users range is odd and which is even and then summing these values. The total sum loop works but I have tried multiple other loops for the odd and even sums and have had no success. So if someone could help me sum these numbers based on whether there even or odd.
#include <iostream>
using namespace std;
int main()
{
int sum = 0, start, endnumber;
int sumall = 0, c=0;
cout << "Please enter the starting integer\n";
cin >> start;
cout << "Please enter the ending integer\n";
cin >> endnumber;
while (endnumber <= start)
{
cout << "Please enter the starting integer\n";
cin >> start;
cout << "Please enter the ending integer\n";
cin >> endnumber;
}
while (start <= endnumber)
{
sumall = sumall + start;
++start;
}
std::cout << "The sum is: " << sumall << std::endl;
return 0;
}
You need to check each number in the range with modulo.
For ex:
if( start%2 == 0 ) // even
{
evenSum+=start;
}
else // Odd
{
oddSum+=start;
}
That will do the job.
You can test for odd numbers by anding with 1.
if ((start & 1) == 0)
evenSum += start;
else
oddSum += start;
Small note when talking to mathematicians: don't use modulo and modulus interchangeably (even though wikipedia says they are the same):
modulo is % for positive numbers.
modulus is the abs function.