How to get an array of random numbers whose sum is 500?

Question

Programming

We need an array of positive random numbers, the sum of the elements of which will be, say, 500. The number of elements is passed as a parameter, of course, but usually there are 3 or 4 of them. No matter
how I try, there are few successes. I tried, roughly speaking, to divide 50 into parts and then multiply them by 10 and another fierce option. Well, the funniest solution is to generate until the required amount is available, poor percent.

Answer 1

Let's assume that the numbers are non-negative integers. If numbers can be zero, do this:
n1=irand(501); // from 0 to 500
n2=irand(501); // from 0 to 500
n3=irand(501); // from 0 to 500
sort(n1,n2,n3); // sort ascending in any suitable way
x0=n1; x1=n2-n1; x3=n3-n2; x4=500-n3;
If numbers can only be positive, then do the same:
n1=irand(497); // from 0 to 496
n2=irand(497); // from 0 to 496
n3=irand(497); // from 0 to 496
sort(n1,n2,n3); // sort ascending in any suitable way
x0=n1+1; x1=n2-n1+1; x3=n3-n2+1; x4=496-n3+1;
The distribution will be slightly uneven (sets containing zeros in the first case and ones in the second will be slightly less common), but this can be corrected by slightly complicating the program.

Answer 2

In your statement of the problem, there are no restrictions on the values ​​of the elements => they can be positive and negative => you don’t need to divide anything, just fill the array with random numbers with a sign, and make one element equal to the difference between the required mum and the sum of the remaining elements.

Answer 3

00Specified range of random number from Ato C
06Remainder B + Awill be the last number

Answer 4

At night, looking, I threw in such a random shit code, but I think you will understand the essence

function random500() {
  var n1 = Math.ceil(Math.random()*499);
  var n2 = Math.ceil(Math.random()*(500-n1));
  var n3 = 500 - n1 - n2;
  console.log(n1 + " " + n2 + " " + n3);
}

You can drive directly in the f12 console. The ceiling is 499 for the first number, respectively, so that 500 does not accidentally come across, then the remaining two (or as many as you need) will be equal to zero. In theory, it should be replaced by 500, but think for yourself.

Answer 5

int getRandom()
{
     return 500;
}

or

//
// код где массив заполняется любыми числами
//
int summ = (array - array) + 500;
printf("Сумма всех чисел в массиве = %d", summ);

or

nachalo:
int massive[10];
int summa = 0;
for(int i = 0; i < 10; i++, array[i] = rand() % 500);
for(int i = 0; i < 10; i++, summa += array[i]);
if(summa != 500)
    goto nachalo;
else if(summa == 500)
    goto konec;
konec: