数的乘积，总和和位数

Question

My aim is to create a program that finds:

• T(P(n)) : The sum of the product of the digits of all numbers up to n. For example, T(p(5))= p(1)+p(2)+p(3)+p(4)+p(5), where p is a function that calculates the product of all nonzero digits in an integer.
• T(S(n)) : Sum of all integers up to n, which is pretty easy using the n*(n+1)/2 formula
• T(D(n)) : Sum of all the digits of all the integers up to n.

I've attempted to complete the first 2 parts, as shown in the codepen link: https://codepen.io/anon/pen/LdjveQ . Javascript shown below:

function result() {
var take = document.getElementById("number").value;
  return eval(take.replace(/(\d)(?=\d)/g, '$1+'));
}
function answer() {
document.getElementById('resultfinal').innerHTML = result();
}

function result2() {
var sake = document.getElementById("integers").value;
  var mult = (sake*(sake+1));
  return eval(mult/2);
}

function answer2() {
document.getElementById('sumofn').innerHTML = result2();
}

I have no idea:

• T(P(n)) : How to approach the aspect of adding the p(n) of all integers greater than 0, but less than n.
• T(D(n)) : Why the n(n+1)/2 formula produces the wrong result
• T(S(n)) : How I should approach T(D(n))

Answer 1

If a "brute force" approach is doable, I would first define P(n) as a function (se p function below) and sum it up using a for :

 function p(inputN) { var n = inputN, m = 1, digit; while (n) { digit = n % 10; if (digit) m *= digit; n = Math.floor(n / 10); } return m; } function result3() { var sum = 0, i, n = document.getElementById("tpn-input").value; for(var i = 1; i <= n; i++) { sum += p(i); } return sum; } function answer3() { document.getElementById('tpn').innerHTML = result3(); } 

 <input type="text" id="tpn-input"><button onClick="answer3()">P(n)</button> <p id = "tpn"></p> 

There are other somewhat more readable approaches using string .split() , .filter() together with .reduce() , but the above will certainly be more performatic.

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And, just for the heck of it, an over-oneliner short implementation:

 function p(n) { return n.toString().split('').map(Number).filter(i => i).reduce((a, b) => a * b, 1); } function tp(n) { return [...Array(+n).keys()].map(i => p(i+1)).reduce((a, b) => a + b, 0); } function answer3() { document.getElementById('tpn').innerHTML = tp(document.getElementById("tpn-input").value); } 

 <input type="text" id="tpn-input"><button onClick="answer3()">P(n)</button> <p id = "tpn"></p> 

Answer 2

You should start by trying using some of the array methods to iterate over an array from 1 to n . reduce would be the most appropriate for this situation: it takes in an array and returns some cumulative value from an operation (or few) done over each element in the array. For example:

 function sumUpToN(n) { return Array.from({ length: n + 1}, (_, i) => i) .reduce((partialSum, thisNum) => partialSum + thisNum); } console.log(sumUpToN(5)); 

Generate your arrays using Array.from like that, and then iterate over them with .reduce . The first argument is the returned value from the previous iteration of .reduce , and the second is the current element in the array. Try using a similar method when you need to iterate over the whole array and come out with a single value on the other end - except instead of plain addition (which will produce the sum of all the elements in the array), use whatever operator(s) are appropriate for the problem in question.