Nabrius 计算出n阶乘中尾部零的个数。 比如11！=39916800， return 2 有没有更省时的解题方法？ long long trailingZeros(long long n) { // write your code here, try to do it without arithmetic operators. unsigned long num=0; while(n>0){ num+=n/5; n=n/5; } return num; }