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Split a numeric array by as

Perhaps the title of the exercise is not very explicit in itself, but the idea is: You must enter
few integers to use, and then when you finish entering all values, which has to be determined been the largest number of all patients admitted to subsequently take each element of the array and divide by that number higher.

Here goes:

 
  # include \u0026lt;iostream>  
 
   using namespace std; 
 
   int size, i, j; 
   double array [20]; 
   double greater; 
 
   int main (void  ) 
 
 {court \u0026lt;\u0026lt;"Please enter how many numbers you wish to use" \u0026lt;\u0026lt;endl; 
 cin>> size; 
  for  (i = 0; i \u0026lt;size; i + +) {
 
 court \u0026lt;\u0026lt;"Please enter data" \u0026lt;\u0026lt;endl; 
 cin>> array [i]; 
 largest = array [i] ; 
   if (array [i]> highest) {
 
 higher = array [i];} 
 
 
 
  } for (j = 0, j \u0026lt;size; j + +) {
 
 array [j] = (array [j] / highest); 
 court \u0026lt;\u0026lt;"The resulting division between the number" \u0026lt;\u0026lt; j +1 \u0026lt;\u0026lt;"and the largest is" \u0026lt;\u0026lt;array [j] \u0026lt;\u0026lt;endl;} 
 
 system ("pause"); 
   return 0;} 
 
 

We used , as in most basic exercises in Dev-C + + , conditional and cycles, and although it seems that most frequently used are used for instruction , also can structure the program to work using the command while, most of all, this is at the convenience of the programmer.

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Serie Fibonacci: Print first 'n' numbers

programming in general, not just C language, use of exercises related to the Fibonacci series is a classic, and in this exercise, the idea is to print the first n numbers series, where n is a predefined value and restrictions of language should not be a very large number.

We will print the first 20 numbers in Dev-C series.

 
  # include \u0026lt;iostream>  
 
   using namespace std; 
 
  int previous1 , anterior2, current, i; 
 
   int main (void  ) 
 
 {court \u0026lt;\u0026lt;"This program will print the first 20 numbers in the series of Fibbonacci "\u0026lt;\u0026lt;endl; 
 system (" pause "); 
 previous1 = 0; 
 anterior2 = 1; 
 court \u0026lt;\u0026lt;previous1 \u0026lt;\u0026lt;endl; 
 court \u0026lt;\u0026lt;anterior2 \u0026lt;\u0026lt;endl ; 
   for (i = 1; i \u0026lt;= 20; i + +) 
 / * The 20 refers to the amount of numbers to print * / 
 
 {+ current = anterior2 previous1, previous1 = anterior2 
; 
 court \u0026lt;\u0026lt;today \u0026lt;\u0026lt;endl; 
 anterior2 = current;} 
 
 system ("pause"); 
   return 0; 
 
} 

Also in this case, it is possible to modify the program so that the user enter the value of 'n', and it would be better instead to report the numbers as int, long double it as , embracing a greater number of digits when the numbers start to become very large, and would read:

 
  previous1  long double, anterior2, current, i; 
 
   long double main (void  ) 
 

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Print odd and even numbers from 1 to 'n'

This is a simple example of the combined use of cycles ( For in this case) and conditionals.
The first idea is to print all the odd numbers from 1 to n, then the couple, noting that "n" is a predefined value, although it could be a simple modification for the user to enter the number.
we go.

 
 \u0026lt;iostream>   # include using namespace 
 
   std; 
 
   int a, b, c, d; 
 
   int main (void) {
 
 court \u0026lt;\u0026lt;"Printing of odd numbers from 1 to 999" \u0026lt;\u0026lt;endl \u0026lt;\u0026lt; ; endl; 
 system ("pause"); 
  for  (a = 1, a \u0026lt;= 1000; a + +) 
 {
 b = a% 2; 
  if  (b == 1) 
 {
 court \u0026lt;\u0026lt;a \u0026lt;\u0026lt;endl;} 
 
 
} system ("pause"); 
 court \u0026lt;\u0026lt;"Print the even numbers from 1 to 1000" \u0026lt;\u0026lt;Endl \u0026lt;\u0026lt;endl; 
 system ("pause"); 
   for (d = 1, d \u0026lt;= 1000; d + +) {
 
 c = d% 2; 
   if (c = = 0) 
 
 {court \u0026lt;\u0026lt;d \u0026lt;\u0026lt;endl;} 
 
 
} system ("pause"); 
   return 0;}