确定阶乘 Function 的尾随零的数量（javascript）

Question

I'm writing a function that returns the number of trailing zeros of a factorial. It seems that the amount of zeros increases from 0, for every 5 positive increments. but also increases 1 zero for every 25 increments, The problem is that when I pass in a number like 1000, it should return 249. but it returns 240 instead? Am I missing something?

I'm almost positive there is a better way to write this, but hopefully, someone can tell me what I'm missing in my current function?

Note: This function must not calculate the actual factorial to determine the number of trailing zeros.

 function zeros(n) { const extraZeros = n/25 const zeros = n/5 if(extraZeros >= 1) { return zeros + Math.floor(extraZeros) }else{ return Math.floor(zeros) } } console.log(zeros(0), 0); console.log(zeros(5), 1); console.log(zeros(6), 1); console.log(zeros(30), 7); console.log(zeros(60), 14); console.log(zeros(1000), 249); // missing 9 zeros?

Any help would be greatly appreciated, Thank you!

Answer 1

It seems that the amount of zeros increases from 0, for every 5 positive increments. but also increases 1 zero for every 25 increments.

You're right, but read this sentence again. Do you see the pattern?

It'll also increase a zero for every 125 and 625 increments, etc. You need to generalize based on powers of 5.

 const zeros = (n) => { let power = 1; let totalZeros = 0; while (n > 5 ** power) { totalZeros += Math.floor(n / (5 ** power)); power++; } return totalZeros; }; console.log(zeros(0), 0); console.log(zeros(5), 1); console.log(zeros(6), 1); console.log(zeros(30), 7); console.log(zeros(60), 14); console.log(zeros(1000), 249);

Another way of writing it, from Patrick Roberts:

 const zeros = (n) => { let totalZeros = 0; for (let power = 5; power < n; power *= 5) { totalZeros += Math.floor(n / power); } return totalZeros; }; console.log(zeros(0), 0); console.log(zeros(5), 1); console.log(zeros(6), 1); console.log(zeros(30), 7); console.log(zeros(60), 14); console.log(zeros(1000), 249);