typedef struct complex {
double real, imag;
} complex;
...
complex x, y, z;
...
/* add */
z.real = x.real + y.real;
z.imag = x.imag + y.imag;
/* sub */
z.real = x.real - y.real;
z.imag = x.imag - y.imag;
/* mul */
z.real = x.real*y.real - x.imag*y.imag;
z.imag =x.imag*y.real + x.real*y.imag;
/* div */
double d = y.real*y.real + y.imag*y.imag;
z.real = (x.real*y.real + x.imag*y.imag)/d;
z.imag = (x.imag*y.real - x.real*y.imag)/d;
To calculate the sum of all even numbers starting from 20 until the sum exceeds 1000, you can initialize a variable for the sum and a counter starting at 20. In a loop, add the counter to the sum and increment the counter by 2 (to keep it even) until the sum exceeds 1000. The final sum will be the total of all even numbers added. Here's a simple pseudocode example: sum = 0 number = 20 while sum <= 1000: sum += number number += 2
public static final int getSum(final int n) { int sum = 0; for(int i = 1; i <= n; ++i) { sum += i; } return sum; }
Sum = Sum + first number Sum = Sum + second number Sum = Sum + third number Average = 1/3 x Sum
if (n%2==0) sum=n/2*(n+1); else sum=(n+1)/2*n;
int sum(int list[], int arraySize) { int sum=0; for(int i=0; i<arraySize; ++i ) sum+=list[i]; return(sum); }
When the sum of a number plus 3 is squared, it is 11 more than the sum of the number plus 2 when squared.
Their sum is real.
13
The sum of a number plus 3 to the square root of 9 to the sum of 2 would equal -2. This is a math problem.
2
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4
2
2
2
2