LeetCode Factorial Trailing Zeroes
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思路:
求阶乘的末尾0的个数。
n的阶乘可以写成:
即所有质因子的乘积,只有一对2和5两个质因子相乘会贡献一个末尾0,又因为x一定大于z,所以只需计算有多少个质因子5即可。
class Solution {public: int trailingZeroes(int n) { int ans = 0; while(n) { ans += n/5; n /= 5; } return ans; }};
0 0
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