leetcode-Factorial Trailing Zeroes
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Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
对n!做质因数分解n!=2x*3y*5z*...
显然0的个数等于min(x,z),并且min(x,z)==z
证明:
对于阶乘而言,也就是1*2*3*...*n
[n/k]代表1~n中能被k整除的个数
那么很显然
[n/2] > [n/5] (左边是逢2增1,右边是逢5增1)
[n/2^2] > [n/5^2](左边是逢4增1,右边是逢25增1)
……
[n/2^p] > [n/5^p](左边是逢2^p增1,右边是逢5^p增1)
随着幂次p的上升,出现2^p的概率会远大于出现5^p的概率。
因此左边的加和一定大于右边的加和,也就是n!质因数分解中,2的次幂一定大于5的次幂
解法一:
从1到n中提取所有的5
class Solution {public: int trailingZeroes(int n) { int ret = 0; for(int i = 1; i <= n; i ++) { int tmp = i; while(tmp%5 == 0) { ret ++; tmp /= 5; } } return ret; }};
解法二:
由上述分析可以看出,起作用的只有被5整除的那些数。能不能只对这些数进行计数呢?
存在这样的规律:[n/k]代表1~n中能被k整除的个数。
因此解法一可以转化为解法二
class Solution {public: int trailingZeroes(int n) { int ret = 0; while(n) { ret += n/5; n /= 5; } return ret; }};
0 0
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