is the solution correct?

3 vues (au cours des 30 derniers jours)
Musaed
Musaed le 12 Mai 2023
Commenté : Walter Roberson le 27 Mai 2023
for numbres 50 70 13 14 7 add all find sum for even find how many odd use one loop
sume=0;
sum=0;
for x=1:5
z = input ('z=')
sum=sum+z;
if mod(z,2)==03
sume=sume+z;
end
if mod(z,2)~=0
count0= count+1;
end
end
is the solution correct?
  1 commentaire
Torsten
Torsten le 12 Mai 2023
Modifié(e) : Torsten le 12 Mai 2023
is the solution correct?
Easy to check by yourself:
Z=[50 70 13 14 7];
sume=0;
sum=0;
for x=1:5
z = Z(x);
sum=sum+z;
if mod(z,2)==03
sume=sume+z;
end
if mod(z,2)~=0
count0= count+1;
end
end
Error using count
Not enough input arguments.

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Réponses (1)

Diwakar Diwakar
Diwakar Diwakar le 27 Mai 2023
Modifié(e) : Diwakar Diwakar le 27 Mai 2023
numbers = [50, 70, 13, 14, 7];
sum_all = 0;
sum_even = 0;
odd_count = 0;
for i = 1:length(numbers)
num = numbers(i);
sum_all = sum_all + num; % Adding to the sum of all numbers
if mod(num, 2) == 0
sum_even = sum_even + num; % Adding to the sum of even numbers
else
odd_count = odd_count + 1; % Counting odd numbers
end
end
fprintf('Sum of all numbers: %d\n', sum_all);
Sum of all numbers: 154
fprintf('Sum of even numbers: %d\n', sum_even);
Sum of even numbers: 134
fprintf('Count of odd numbers: %d\n', odd_count);
Count of odd numbers: 2
output:
Sum of all numbers: 154
Sum of even numbers: 134
Count of odd numbers: 3
  1 commentaire
Walter Roberson
Walter Roberson le 27 Mai 2023
We recommend against providing complete solutions to homework problems.

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