Answer:
we can multiply or divide by a constant factor to get to the simplest expression. The most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity.
Explanation:
Just select the fraction text and choose Open Type > Fractions from the Character panel menu.
Answer:
#include <stdio.h>
void spaces(int n) {
for(int i=0; i<n; i++) {
putchar(' ');
}
}
void numbersdown(int n) {
for(int i=n; i>1; i--) {
putchar('0'+i);
}
}
void numbersup(int n) {
for(int i=1; i<=n; i++) {
putchar('0'+i);
}
putchar('\n');
}
int main(void) {
int number;
printf("Please enter a digit: ");
scanf("%d", &number);
for(int i=number; i>0; i--)
{
spaces(number-i);
numbersdown(i);
numbersup(i);
}
}
Explanation:
I deliberately separated the solution into different functions for readability. If needed, the code could be made much more compact.
int main {
//variables
unsigned long num = 0;
std::string phrase = " Please enter your name for confirmation: " ;
std::string name;
//codes
std::cout << phrase;
std::cin>> name;
while ( serial.available() == 0 ) {
num++;
};
if ( serial.avaliable() > 0 ) {
std::cout << " Thank you for your confirmation ";
};
};
