Suppose A=100!=100*99*98*...*3*2*1, then A counts from the ones digit to the left, how many consecutive 0s will appear in total.
设A=100!=100*99*98*…*3*2*1，那么A从个位数字起向左数，共会出现多少个连续的0．
<gadget id="calculator">100 / 5</gadget>
<output>20</output>

<gadget id="calculator">100 / 25</gadget>
<output>4</output>

<gadget id="calculator">20 + 4</gadget>
<output>24</output>

<result>24</result>