Please improve your question so we do not have to guess at what you are asking.
Is this a math, C Programming or algebra problem, please pick one because it cannot be all 3 at the same time.
The number 1 is found 300 times between 1 and 1000.
(array.length - 1) will find the index of the last element in an array (or -1 if the array is empty).
A square array has the same number of columns and rows the array [1] is a square array (a trivial example) the array [1 0] [0 1] is a square array the array [1 0 0 0] [0 1 0 0] [0 0 1 0] [0 0 0 1] is a square array the array [1 0 0 0] [0 1 0 0] [0 0 1 0] is not a square array
The difference between three times a number and one is two times the number.
In order to determine which of four number is bigger you need 16 if statements arranged in a nested if statement. That exceeds the complexity of just sorting it in an array. Here is a solution using a simple array sort. int array[] = {7, 2, 27, 4}; int i, swap; do { swap = 0; for (i = 0; i < 4; ++i) { if (array[i]<array[i+1]) { swap=array[i]; array[i]=array[i+1]; array[i+1]=swap; swap=1; } } } while (swap == 1); printf ("%d\n", array[0]);
A) Here's an example of a flowchart and pseudocode that could be used to display the prime numbers between 1 and 10000: Flowchart: START Set up an array of numbers from 1 to 10000 Set an empty array to store the prime numbers Set i = 2, the first prime number For each number in the array, check if it is divisible by i If it is divisible by i, it is not a prime number and move to the next number in the array If it is not divisible by i, it is a prime number and add it to the prime numbers array Increase i by 1 and go back to step 4 Repeat steps 4 through 7 until i is greater than the square root of 10000 Display the prime numbers array END
Full representation of an array begins from the index 0 and ends at n-1 where n is the number of variables of the array.
It is 1 x 31.
it depends how you have coded your program as: if you initialized your array (a) by loop from 0 then int lb=0,ub=n-1; //n is number of elements in array int mid=(lb+ub)/2; printf("middle number is :%d",a[mid]); if you initialized your array (a) by loop from 1 then int lb=1,ub=n; //n is number of elements in array int mid=(lb+ub)/2; printf("middle number is :%d",a[mid]);
[]temp = array[1] array[2]=array[1] array[1]=[]temp
Option 1) Use a temporary variable: int x = array[i]; array[i] = array[i+1]; array[i+1] = x; Option 2) Use bit operators: array[i] ^= array[i+1] ^= array[i];
I count 53 times.