Description

Write this function

for i from 1 to n, do i % m and return the sum.

f(n=10, m=5) // returns 20 (1+2+3+4+0 + 1+2+3+4+0)

You'll need to get a little clever with performance, since n can be a very large number

Kata's link: Sum of many ints

Best Practices

JS First:

f = function(n, m) {
  sum1toN = function(x) {
    return x * (x + 1) / 2;
  };

  return (sum1toN(m - 1)) * (Math.floor(n / m)) + sum1toN(n % m);
};

JS Second:

function f(n, m) {
  return Math.floor(n / m) * m * (m - 1) / 2 + (n % m) * (n % m + 1) / 2
}

JS Third:

function f(n, m) {
  return gauss(m - 1) * (n / m | 0) + gauss(n % m)
}

function gauss(n) {
  return n * (n + 1) / 2
}

JS Fourth:

function f(n, m) {
  return Math.floor(n/m) * (m-1)*m/2 + (n%m)*(n%m+1)/2;
}

JS Fifth:

function f(n, m) {
  return ~~(n/m)*m*(m-1)/2 + n%m * (n%m + 1)/2;
}