CODEWITHSUNDEEP
javascriptarraysstringnumber-formatting
Sarvar
Sarvar

Reputation: 27

Problems with calculating the sum of powers

Here is the task.

Given an array, a machine outputs the sum of every other digit to the power of the next digit.
For example:
Given array [w, x, y, z], the output would be the value of

Math.pow( w, x ) + Math.pow( y, z ).

QUESTION

Find the output of the machine given array: [98, 45, 97, 36, 22, 62, 88, 71, 16, 20, 54, 59, 23, 31, 12, 23, 77, 39, 37, 51, 68, 69, 92, 30].

Here is my code written in JS which is giving the wrong result I am getting the result of

1143588350561521212653379541203320722577849507571892512032437843441247325541362219974575744439494976388658582305054383575339973680209682418

which is wrong. 

let a = [98, 45, 97, 36, 22, 62, 88, 71, 16, 20, 54, 59, 23, 31, 12, 23, 77, 39, 37, 51, 68, 69, 92, 30];
let s = 0,
  r = 1;
for (let i = 0; i <= 22; i = i + 2) {
  for (let j = 1; j <= a[i + 1]; j++) {
    r = multiply(r, a[i]);
  }
  s = sum(s, r);
  r = 1;
}

// Add big numbers as strings in order to avoid scientific (exponent) notation
function sum(arg1, arg2) {
  var sum = "";
  var r = 0;
  var a1, a2, i;

  if (arg1.length < arg2.length) {
    a1 = arg1;
    a2 = arg2;
  } else {
    a1 = arg2;
    a2 = arg1;
  }
  a1 = a1.toString().split("").reverse();
  a2 = a2.toString().split("").reverse();

  for (i = 0; i < a2.length; i++) {
    var t = ((i < a1.length) ? parseInt(a1[i]) : 0) + parseInt(a2[i]) + r;
    sum += t % 10;
    r = t < 10 ? 0 : Math.floor(t / 10);
  }

  if (r > 0)
    sum += r;
  sum = sum.split("").reverse();

  while (sum[0] == "0")
    sum.shift();

  return sum.length > 0 ? sum.join("") : Number("");
}


// Multiply big numbers as strings in order to avoid scientific (exponent) notation
function multiply(a, b) {
  var aa = a.toString().split('').reverse();
  var bb = b.toString().split('').reverse();

  var stack = [];

  for (var i = 0; i < aa.length; i++) {
    for (var j = 0; j < bb.length; j++) {
      var m = aa[i] * bb[j];
      stack[i + j] = (stack[i + j]) ? stack[i + j] + m : m;
    }
  }
  for (var i = 0; i < stack.length; i++) {
    var num = stack[i] % 10;
    var move = Math.floor(stack[i] / 10);
    stack[i] = num;

    if (stack[i + 1])
      stack[i + 1] += move;
    else if (move != 0)
      stack[i + 1] = move;
  }
  return stack.reverse().join('');
}
// Print the result
console.log(s);

Upvotes: 0

Views: 483

Answers (2)

fafl
fafl

Reputation: 7385

As Dominic said, use any big integer library:

function calc(arr) {
  result = bigInt();
  for (var i = 0; i < arr.length; i += 2) {
    result = result.plus(bigInt(arr[i]).pow(arr[i+1]))
  }
  return result.toArray(10).value.join('');
}

console.log(calc([98, 45, 97, 36, 22, 62, 88, 71, 16, 20, 54, 59, 23, 31, 12, 23, 77, 39, 37, 51, 68, 69, 92, 30]))
<script src="https://peterolson.github.io/BigInteger.js/BigInteger.min.js"></script>

Javascript can not handle big integers natively without losing precision.

Upvotes: 0

Dominic Price
Dominic Price

Reputation: 1146

I'd recommend using the big-integer library:

let a = ...;
let s = bigInt()
for (var i = 0; i < a.length; i += 2) 
  s = s.plus(bigInt(a[i]).pow(bigInt(a[i+1])));

Upvotes: 2

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