Response:
#code (count_seq.py)
def count_seq():
n='2'
while True:
yield int(n)
next_value=''
while len(n)>0:
first=n[0]
count=0
while len(n)>0 and n[0]==first:
count+=1
n=n[1:]
next_value+='{}{}'.format(count,first)
n=next_value
if __name__ == '__main__':
gen=count_seq()
for i in range(10):
print(next(gen))
Clarification:
- Begin with the number 2. Utilize a string for easier manipulation rather than integers.
- Engage in an infinite loop.
- Yield the current integer value of n.
- Continue looping until n becomes an empty string.
- Repeat as long as n has content and the first digit matches the leading digit.
- Concatenate the count and the first digit to form next_value.
Answer:
Below is the explanation for the C code.
Explanation:
#include <stdio.h>
#include <stdbool.h>
int main(void) {
int userNum;
bool isPositive;
bool isEven;
scanf("%d", &userNum);
isPositive = (userNum > 0);
isEven = ((userNum % 2) == 0);
if(isPositive && isEven){
printf("Positive even number");
}
else if(isPositive &&!isEven){
printf("Positive number");
}
else{
printf("Not a positive number");
}
printf("\n");
return 0;
}
Answer:
SyntaxError.
Explanation:
https://www.quora.com/In-Python-what-kind-of-error-is-returned-by-the-following-code-e-g-NameError-ValueError-IOError-etc-def-my_func-n1-n2-return-n1-n2-my_func-1-2-3 original source
brainlest plz
Respuesta: Te proporcioné 6 opciones de las cuales puedes elegir.
Integridad
Escalabilidad
Calidad de Servicio
Tolerancia a Fallos
Redes de Línea Eléctrica
Seguridad
This problem is a personal reflection and cannot be solved without discussing it after the individual provides input.
