Respuesta :
Answer:
ALGORITHM READ_NUMBER()
1. SET NUM := 0 [INITIALLY SET THE VALUE OF NUM TO 0]
2. WRITE : “Input a Number” [DISPLAY A MESSAGETO USER]
3. READ NUM [STORE THE VALUE TO NUM]
4. IF NUM<0 [IF VALUE OF NUM IS -VE THEN MULTIPLY IT BY -1]
5. NUM := NUM * (-1)
6. [END OF IF]
7. RETURN NUM [RETURN THE NUMBER]
1. ALGORITHM DIGIT_COUNT(NUM)
SET COUNT := 0
SET R:=0
REPEAT WHILE(NUM != 0)
R := NUM MOD 10
COUNT : = COUNT+1
NUM := NUM/10
[END OF LOOP]
RETURN COUNT
2. ALGORITHM DIGIT_ODDCOUNT(NUM)
SET COUNT := 0
SET R:=0
REPEAT WHILE(NUM != 0)
R := NUM MOD 10
IF (R MOD 2 != 0)
COUNT : = COUNT+1
[END OF IF]
NUM := NUM/10
[END OF LOOP]
RETURN COUNT
3. ALGORITHM SUM_OF_DIGITS(NUM)
SET SUM := 0
SET R:=0
REPEAT WHILE(NUM != 0)
R := NUM MOD 10
SUM : = SUM+R
NUM := NUM/10
[END OF LOOP]
RETURN SUM
C++ PROGRAM
#include<iostream>
using namespace std;
//method to count the number of digits
int digit_count(int num)
{
int count=0; //set count to 0
int r;
//loop will continue till number !=0
while(num!=0)
{
r=num%10; //find the individual digits
count++; //increase the count
num=num/10; //reduce the number
}
return count; //return the count
}
int odd_digit_count(int num)
{
int count=0;//set count to 0
int r;
//loop will continue till number !=0
while(num!=0)
{
r=num%10;//find the individual digits
if(r%2!=0) //check for odd digits
count++;//increase the count
num=num/10;//reduce the number
}
return count;//return the count
}
int sum_of_digits(int num)
{
int sum=0;
int r;
//loop will continue till number !=0
while(num!=0)
{
r=num%10;//find the individual digits
sum=sum+r;//find the sum of digits
num=num/10;//reduce the number
}
return sum;//return sum
}
//input method
int input()
{
int x;
cout<<endl<<"Enter a number";
cin>>x;//read the number
if(x<0) //if -ve then turn it to +ve
x=x*-1;
return x; //return the number
}
//driver program
int main()
{
int n,dc,odc,sod;
n=input(); //call to input()
dc=digit_count(n); //call to digit_count()
odc=odd_digit_count(n); //call to odd_digit_count()
sod=sum_of_digits(n);//call to sum_of_digits()
//display the details
cout<<endl<<"The positive number is : "<<n;
cout<<endl<<"The number of digits in "<<n<<" is "<<dc;
cout<<endl<<"The number of odd digits in "<<n<<" is "<<odc;
cout<<endl<<"The sum of digits of "<<n<<" is "<<sod;
}
Explanation: