Hi can anyone help me in creating a code for the sum of input number in C++.
e.g.
Input: 535
Output: 13
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int i = 0,sum = 0;
int numarray[50];
void calc(int c,string num) {
do {
numarray[i] = num.at(i);
/*sum = sum + numarray[i];*/
sum = sum + num.at(i);
cout << numarray[i] << endl;
i++;
} while(i != c);
cout << sum << endl;
}
int main() {
string num;
int i,charlen;
int numar[50];
int sum=0;
cout << "Input numbers: ";
cin >> num;
charlen = num.length();
calc(charlen,num);
}
While characters are numbers, their values are often not the same as the character they represent. The most common encoding scheme is ASCII encoding.
As you can see in the linked table, the digits don't have the value 1 or 2 etc. Instead they have values like 49 and 50 etc. The characters '5' and '3' have the ASCII values 53 and 51 respectively, adding e.g. "535" will give you the result 157.
But as you can see in the ASCII table, all numbers are consecutive, that means we can use a very simple trick to get the digits value from its ASCII value, simply subtract the ASCII value of '0'. For example '5' - '0' will give you the value 5.
Without testing:
#include <iostream>
int main()
{
unsigned long long x;
std::cout << "Enter a non-negative number: ";
std::cin >> x;
unsigned long long sum = 0;
do { sum += x % 10; } while ( x /= 10 );
std::cout << "\nThe sum of digits of the number is " << sum << std::endl;
}
For the sake of readability, it might help to not pass in the charlength into the function, and just do that inside the method. Besides that, a for loop seems like a better option in this case as well.
void calc(string num) {
int sum = 0, int i = 0;
//for loop run from num.length - 1 to 0 {
/*numarray[i] = stoi(num.at(i));*/
/*sum = sum + numarray[i];*/
sum = sum + stoi(num.at(i));
cout << numarray[i] << endl;
i++;
}
cout << sum << endl;
}
Related
Total Noob here, I am having a hard time with an assignment. I am taking a beginner course in C++ and have to figure out how to calculate the sum of negative integers and their avg. Sum of positive integers and the avg. And the sum of all numbers and the avg. I have gotten the last part already but how do I calculate the sum of negative integers and avg, and positive integers and avg using a while loop?
I provided my code below.
#include <iostream>
using namespace std;
#include <iomanip>
int main(int argc, const char * argv[]) {
int x;
double avg = 0.0;
int count = 0;
int sum = 0;
// ask users for input
cout << ("Welcome to the greatest calculator!\n");
cout << ("Please enter 10 integers seperated by spaces \n");
do {
std::cin >> x;
sum = sum + x;
count = count + 1;
}
while (count < 10);
// calculate average
avg = sum/10.0;
// output average
cout << fixed;
cout << "For all 10 numbers the sum is " << sum << "." "The average is " << setprecision (2) << sum/10.0 <<".\n";
return 0;
}
The output should look something like this.
Please enter 10 integers separated by spaces:
1 -1 45 17 28 -2 0 9 -14 11
Upon our intelligent calculations, here is the result:
+ There are 7 positive numbers, sum = 111.00 and average = 15.86
+ There are 3 negative numbers, sum = -17.00 and average = -5.67
+ For all 10 numbers, sum = 94.00 and average = 9.40 */
Use two variable int negativeVar=0 , PositiveVar=0 . In the loop try a condition if(GivenNumber<0) to detect the given number is negative or positive. Then add all positive and negative value separately and make avarage.
(Sorry for bad english)
You can do like this (notice comments):
#include <iostream>
int main(void) {
// Declaration and initialization of the required variables
float cPositive = 0.0f;
float cNegative = 0.0f;
int it = 0;
std::cout << "Enter 10 numbers (floating point assignable): \n";
// Looping till 10 iterations
do {
float temp;
std::cin >> temp;
// If the number is greater than zero, i.e. (+ve) then cPositive sums up
// otherwise, cNegative
if (temp > 0) cPositive += temp;
else if (temp <= 0.0f) cNegative -= temp;
} while (++it < 10); // Increment and comparison together
// Final results
std::cout \
<< "Sum of positive: " << cPositive << std::endl
<< "Sum of negative: -" << cNegative << std::endl;
return 0;
}
A simple test case:
Enter 10 numbers (floating point assignable):
10.5
-1.5
2.2
5.5
-3.8
-99.3
10
4.5
-1.0
0
Sum of positive: 32.7
Sum of negative: -105.6
Moreover, if you want to see average, then declare two variables, pos and neg where both are initially zero. After that, when a positive number or negative number occurs, just increment pos or neg and divide with them by cPositive or cNegative respectively.
#include <iostream>
#include <string>
using namespace std;
int main()
{
// lets declare some variable first.
int positiveSum =0; //this will hold sum of positive nums
int negativeSum =0; // this will hold sum of negative nums
int totalSum =0; // this will hold sum of all the nums
int number=0; // user input for number
for (int i = 1; i <=10; i++) // loop from 1 to 10 times
{
cout << " Enter a number: ";
cin >> number;
// now check if number is positive or negative
if (number >=0)
{
positiveSum += number; // adds this number to positiveSum
}
else if (number < 0)
{
negativeSum += number; // adds this number to negativeSum
}
}
// So finally add the positiveSum and negativeSum to get the totalSum
totalSum = positiveSum + negativeSum;
cout << endl;
cout << " Total of Positive numbers is: " << positiveSum << endl;
cout << " Total of Negative numbers is: " << negativeSum << endl;
cout << " Total of all numbers is: " << totalSum << endl;
return 0;
}
The code below produces the following output:
$ ./main
The (sum, avg) of negative integers = (-15, -5)
The (sum, avg) of positive integers = (6, 2)
The (sum, avg) of all numbers = (-9, -1.5)
Please read the comments because they are in fact the detailed answer.
#include <array>
#include <iostream>
int main()
{
// For convenience, keep the numbers in an std::array. std::vector is
// equally convenient.
std::array<int, 6> integers { 1, -4, 2, -5, 3, -6 };
// Define variables that store the sums and the counts.
int positiveSum = 0;
int positiveCnt = 0;
int negativeSum = 0;
int negativeCnt = 0;
// Iterate over the numbers taking one of them at a time.
int i = 0;
while (i < integers.size())
{
int number = integers[i];
// Is the number positive?...
if (number >= 0)
{
// ... it is - add it to the positive sum and increment the count.
positiveSum += number;
++positiveCnt;
}
// The number is not positive, so it must be negative...
else
{
// ... add it to the negative sum and increment the count.
negativeSum += number;
++negativeCnt;
}
// Get ready for the next number.
++i;
}
// Time to print out the results.
// Note that before we calculate the average, we have to cast at least one
// of the terms of the division to floating point type. Otherwise the
// division will be done with integers where the result is also an integer
// (e.g. 3 / 2 -> 1).
// Only affter the casting you will be getting expected answers
// (e.g. double(3) / 2 -> 1.5).
std::cout <<
"The (sum, avg) of negative integers = (" <<
negativeSum << ", " <<
double(negativeSum) / negativeCnt << ")" << std::endl;
std::cout <<
"The (sum, avg) of positive integers = (" <<
positiveSum << ", " <<
double(positiveSum) / positiveCnt << ")" << std::endl;
std::cout <<
"The (sum, avg) of all numbers = (" <<
negativeSum + positiveSum << ", " <<
double(negativeSum + positiveSum) / (negativeCnt + positiveCnt) << ")" << std::endl;
}
#include <iostream>
using namespace std;
int main()
{
char op;
float num1,num2;
cout << "Enter two operands: ";
cin >> num1 >> num2;
switch(op)
{
case '+':
cout << num1+num2;
break;
case '-':
cout << num1-num2;
break;
case '*':
cout << num1*num2;
break;
case '/':
cout << num1/num2;
break;
default:
//If the operator is other than +,-,*,/, error message is shown.
cout << "Error! operator is not correct";
break;
}
return 0;
}
I'm starting learning C++ on my own and I'm confused with one assignment which I'm trying to complete. User shoud type in natural numbers as long as he wants until he types in 0. After that my program should find the largest sum of digits which were typed and print it out. It shoud also print out a number from which it took the sum.
Here is what I tried to do:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int input = 0;
int digit;
int sum = 0;
int largest = 0;
do
{
cout << "enter a natural number (0 if done): " << flush;
cin >> input;
while (input > 0)
{
digit = input % 10;
sum = sum + digit;
input = input / 10;
}
if (sum > largest)
largest = sum;
} while (input);
cout << "Max sum of digits was " << largest << "for" << endl;
}
When I'm running the programm it counts sum of digits from only first typed in number and stop working. When I take while (input > 0) away, it makes a loop, but doesn't count digits.
I'll be very grateful for help and explanation.
P.S. Sorry for my English, I'm not a native speaker.
You seem to have three problems here:
1 - You are trying to use a variable that you essentially set to zero in your while loop
2 - You seem to be looking for the input that is for the largest sum
3 - You are not resetting your sum variable for each input
The solution for the first problem is to "backup" the input into another variable before modifying it and using that variable for the while loop.
That also allows for you to get what the largest number inputted is and store it.
int input = 0;
int inputBackup = 0;
int digit;
int sum = 0;
int largest = 0;
int largestInput = 0;
To add in the inputBackup variable, put it after the cin.
Then set the largestInput in your sum > largest if statement to set the largestInput if it is the largest.
cout << "enter a natural number (0 if done): " << flush;
cin >> input;
inputBackup = input;// This line
sum = 0; // and this line
while (input > 0)
{
digit = input % 10;
sum = sum + digit;
input = input / 10;
}
if (sum > largest)
{
largest = sum;
largestInput = inputBackup;// Store largest input
}
Then change while(input) to while(inputBackup) to check the inputBackup variable instead of the input one.
Change your cout to be like this to add the largestInput variable into it
cout << "Max sum of digits was " << largest << " for " << largestInput << endl;
And your code should be fixed!
Happy Coding!
do
{
cout << "enter a natural number (0 if done): " << flush;
cin >> input;
//more code
} while (input);
To make this work correctly, input may not change between the cin and the loop condition.
But
while (input > 0) {
digit = input % 10;
sum = sum + digit;
input = input / 10;
}
does change input.
Replace it with something like
int input2 = input;
while (input2 > 0) {
digit = input2 % 10;
sum = sum + digit;
input2 = input2 / 10;
}
In this part:
while (input > 0) {
digit = input % 10;
sum = sum + digit;
input = input / 10;
}
while input is not zero it'll repeat, so when get out the loop the value of input is 0. Use a auxiliary variable or enclose this code on a function:
int getDigitsSum(int input) {
while (input > 0) {
digit = input % 10;
sum = sum + digit;
input = input / 10;
}
return sum;
}
Try that instead
If we do not zero sum value, it accumulate sum of all input digits and the sum will always be larger than than largest value, bacause it stores largest + sum of current values digits. So, if we zero sum value it only contains sum of digits of current input and can be simple compared with previous one wich was largest.
#include <iostream>
#include <conio.h>
using namespace std;
int main() {
int input = 0;
int digit;
int sum = 0;
int largest = 0;
do
{
while (input > 0) {
digit = input % 10;
sum = sum + digit;
input = input / 10;
}
if (sum > largest)
largest = sum;
sum = 0; // set to 0 current sum
cout << "enter a natural number (0 if done): " << flush;
cin >> input;
} while (input);
cout << "Max sum of digits was " << largest << " for" << endl;
_getch();
return 0;
}
I have with my code. This is about recursion. I have to create function digitAppear( int findDigit, int value) where value is the user input, and findDigit is single digit number ranging from 0 to 9. The function read user input and return each digit number from the user input and count how many times each digit number occurs in the user input. For example, if I type 1234 then the output say 1 appear 1 time, 2 appear 1 time and so on (I hope my explanation is clear) The problem is the only run once and can only return 1 value.
#include <iostream>
using namespace std;
int countOccurence(int, int);
int main()
{
int findDig;
int value;
int n = 0;
cout << "Please enter a positive number: " << endl;
cin >> value;
cout << "The value is " << value << endl;
while ((value < 0) || (value > 9999))
{
cout << "Invalid value. Please try again!" << endl;
cout << "Please enter a positive number: " << endl;
cin >> value; //you need this here, otherwise you're going to be stuck in an infinite loop after the first invalid entry
}
//process the value
for (findDig = 0; findDig < 9; findDig++)
{
cout << endl;
cout << cout << "the " << findDig << "appear in digit " << value << " is " << countOccurence(findDig, value) << " times" << endl;
}
//countOccurance(findDig, value);
//cout
}
int countOccurence(int findDig, int value)
{
int n = value;
while( n > 10 )
{
int a = n / 10; //eliminate the right most integer from the rest
int aa = n % 10; //separate the right most integer from the rest
int b = a / 10; //eliminate the second integer from the rest
int bb = a % 10; //separate the second integer from the rest
int c = b / 10; // eliminate the third integer from the rest
int cc = b % 10; //separate the third integer from the rest
for (findDig = 0; findDig < 9; findDig++)
{
int i = 0;
if (findDig == aa) // see if the findDigit value is equal to single digit of b;
{
i += 1;
} else
{
i += 0;
}
return i;
if (findDig == bb)
{
i += 1;
} else
{
i += 0;
}
return i;
if (findDig == cc)
{
i += 1;
} else
{
i += 0;
}
return il;
}
}
}
The problem is my function countOccurence() doesn't seems right. I wonder if there a way to do it. I have been stuck with this for days and I really appreciate your input, thank you.
To use recursion, you must think about the problem in a different way.
The easiest way of thinking about how you could incorporate recursion into the function is the process of 'peeling off' each number.
A very simple way of doing this is by looking at the first/last digit in the number, compute that, then call itself on the remainder of the number.
Hopefully you can figure out the code from there.
If you mean that function digitAppear itself has to be recursive then it can look the following way as it is shown in the demonstrative program below
#include <iostream>
#include <cstdlib>
size_t digitAppear( int findDigit, int value )
{
return ( std::abs( value ) % 10u == std::abs( findDigit ) ) +
( ( value /= 10 ) ? digitAppear( findDigit, value ) : 0 );
}
int main()
{
int i = 0;
for ( int x : { 0, 11111, 1234, 34343 } )
{
std::cout << "There are " << digitAppear( i, x )
<< " digit " << i
<< " in number " << x
<< std::endl;
++i;
}
return 0;
}
The program output is
There are 1 digit 0 in number 0
There are 5 digit 1 in number 11111
There are 1 digit 2 in number 1234
There are 3 digit 3 in number 34343
Of course you may rewrite function main as you like for example that it would count each digit in a number.
I can get the sum every time the user inputs an integer until either a negative number or non-integer is inputted. Problem is my sum calculations are off. I.E user putting 1000; sum outputs 1111, then user inputs 2000, it adds up to 3333. Just any advice is appreciated. I'll still experiment around with my coding.
#include <iostream>
#include <cstdlib>
using namespace std;
int main()
{
int j , i = 0, k = 0,number;
double sum = 0;
cout << "Enter Positive integer number: ";
while(cin >> number)
{
cout << endl;
if( number < 0)//test if the number is negative
{
cout << "Ending program since user has input a negative number" <<endl;
break;
}
int temp = number;
int p = 1;
while( temp > 0) //counting number of digits
{
sum = sum+temp; //Sum attempt.
temp /= 10;
p *= 10;
i++;
}
cout << sum << endl;
j = i % 3;
p /= 10;
while( i > 0 )//display integer number with 1000 seperator
{
//this is giving me error
cout << char ((number/p) +'0');
number %= p;
p /= 10;
i--;
k++;
j--;
if ((k % 3 == 0 && i > 0)||(j == 0 && i > 2) )
{
cout <<",";
k = 0;
}
}
cout << endl << endl;
cout << "This program will exit if you input any non-integer characters\n";
cout << "Enter another integer number: ";
}
return 0;
}
It looks like you're trying to output an integer number with commas inserted at 1000 boundaries. ie: 1000000 would be displayed as 1,000,000.
This being the case, the easiest way to approach it might not be involving maths but simply to get a string representation of the int (atoi() for example) and count through that. From the back, count forward three chars, insert a comma, repeat until you run out of string.
The specifics of string handling are left as an exercise for the reader - looks like it's his homework after all. ;-)
I'd like to request some help on my HW. I think I'm really close to figuring this out. Our CompSci class is currently shifting from learning Python to (introductory) C++. Since the two are vaguely similar, we've been advised, since we're beginners, to code the problem in Python (which we're very familiar with) and to translate it into C++ using the basics we just learned. The problem to solve is a simple "add the consecutive integers from 1 to that number, given a positive integer input." So an example would be:
>>Enter a positive integer: 10
>>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
The Python code (this was successful) that I'm attempting to translate into C++ is:
num = int(raw_input("Enter a positive integer: "))
sum = 0
for i in range(1, num):
sum += i
print i, "+",
print num, "=", sum+num
And my unsuccessful C++ code:
#include <iostream>
using namespace std;
int main()
{
int num;
int sum;
int i;
sum = 0;
cout << "Please enter a positive integer: " << endl;
cin >> num;
for (i=0; 1 <= num; i++)
{
sum = sum + i;
cout << i << "+" << endl;
}
cout << num << "=" << sum + num << endl;
return 0;
}
But the output is simply an infinite, non-ending addition sequence from 0 to infinity, going top to bottom. Even worse is that it did not print in a straight line like I want it. As you can see, I quite literally tried to translate it word-for-word; I thought that'd be foolproof. Something must be wrong with my for loop. Since C++ doesn't have a class of its own for "range" like Python does, I thought the middle condition statement ("1 <= num;") would act as the range. Why didn't my "=" sign print out? And I don't understand why it won't terminate when it reaches "num." Think you can help? I thank you in advance for the replies.
Fixed code:
#include <iostream>
using namespace std;
int main()
{
int num;
int sum;
int i;
sum = 0;
cout << "Please enter a positive integer: " << endl;
cin >> num;
// Here you had 1 <= num which was always true for positive num
// and it did not depend on value of i.
for (i = 1; i < num; ++i)
{
sum = sum + i;
cout << i << "+"; // Here you had endl which produced newline characters.
}
cout << num << "=" << sum + num << endl;-
return 0;
}
This:
for (i=0; 1 <= num; i++)
should be:
for (i=0; i <= num; i++)
try this.
#include <iostream>
using namespace std;
int main()
{
int num;
int sum;
int i;
sum = 0;
cout << "Please enter a positive integer: ";
cin >> num;
for (i=0; i < num; i++)
{
sum = sum + i;
cout << i << " + ";
}
cout <<num << " = " << sum+num << endl;
return 0;
}
I don't really know Python, but the code
for i in range(1, num):
looks really similar to
for (int i=1; i <= num; ++i)
or is it possibly
for (int i=1; i != num; ++i)
which looks more like C++?
loop in c++ are most basic than python, the for loop is more simpler, it is based on the three expression: initializer expression, the loop test expression, and the counting expression. In particular what is wrong in your code is the test expression. Remember that the loop is executed if the test expression is true. You need to loop if the condition i<num is true. Your loop is never ending because num is always >= 1, or as you wrote 1 <= num always.
To print everythig on a line don't use endl