问题描述:
Write a program that will calculate the number of trailing zeros in a factorial of a given number.
N! = 1 * 2 * 3 * ... * N
Be careful 1000!
has 2568 digits...
For more info, see: http://mathworld.wolfram.com/Factorial.html
Examples
zeros(6) = 1
# 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720 --> 1 trailing zero
zeros(12) = 2
# 12! = 479001600 --> 2 trailing zeros
Hint: You're not meant to calculate the factorial. Find another way to find the number of zeros.
刚刚开始做的时候,很直接的思路就是先求阶乘,然后求尾部0的个数。后来发现,当数据超过一定位数时,js会以科学计数法表示,不能直接计算0的个数。
接着就是思考有2*5能出来0,然后5的个数相对于2的个数会比较少,所以就求5作为因子出现了多少次。
我的答案:
function zeros (n) { var num=0; for(var i=1;i<=n;i++){ var j=i; while(j%5==0){ num=num+1; j=j/5; } } return num; }
优秀答案:
1 function zeros (n) { 2 var zs = 0; 3 while(n>0){ 4 n=Math.floor(n/5); 5 zs+=n 6 } 7 return zs; 8 }