刚看到这道题时没多考虑,就是利用阶乘算出结果然后转为字符数组,判断零的个数(刚发现判断的地方写错了),但是如果输入的值非常大,超过了取值范围,就会报错,而且这道题的思路不是这样的(我是个没有脑子的人!)
思路:乘积中出现0一定是2和5的乘积具体思路就不写了,别人都写好了
参考地址:https://blog.csdn.net/wutingyehe/article/details/46882181
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public class Second {
public static void main(String[] args) {
// TODO Auto-generated method stub
long res = trailingZeros(105);
System.out.println(res);
}
public static long trailingZeros(long n)
{
long Num = jiecheng(n);
String Numst = String.valueOf(Num);
char [] ch = Numst.toCharArray();
long count = 0;
for (int j =0;j<ch.length;j++)
{
if(ch[j]=='0')
{
count++;
}
}
return count;
}
public static long jiecheng(long n)
{
long num =1;
for (int i =1;i<=n;i++)
{
int tmp = i;
num = num* tmp;
}
return num;
}
}