#include <iostream>
#include <vector>
std::vector<int> v;
int main(int argc, char* argv[]) {
while(1) {
int temp;
std::cout << "\nEnter a number: ";std::cin>>temp;
if(temp<0) {
std::cout << "\nEven number(s) is/are:\n---------------------\n";
for(int i=0;i<v.size();i++) {
if(v.at(i)%2==0) std::cout << v[i] << " ";
else continue;
}
std::cout << std::endl;
break;
}else {
v.push_back(temp);
}
}
return 0;
}
Using the knowledge in computational language in python it is possible to write a code that computes an integer's checksum.
<h3>Writting the code:</h3><em>def checkSum(n): # Function definition</em>
<em> total = 0 # total variable which returns sum</em>
<em> c = 0 # counter variable which count the even position</em>
<em> while(n!=0): # running a loop while n != 0</em>
<em> r = int(n%10) #breaking the number into digits</em>
<em> c=c+1 # increase counter</em>
<em> if c%2==0: # if counter is even position then double the number </em>
<em> r = r*2</em>
<em> while(r!=0): # do the process</em>
<em> total = total + int(r%10)</em>
<em> r = int(r/10)</em>
<em> n=int(n/10) </em>
<em> else: # if counter is odd position then simple add the digit </em>
<em> total = total + r # into total</em>
<em> n=int(n/10)</em>
<em> return total # in the last return the sum of digits</em>
<em># Driver code</em>
<em>n = int(input("Enter Number:")) # asking user to enter the number </em>
<em>print(checkSum(n)) # calling the function</em>
See more about python at brainly.com/question/18502436
#SPJ1
