172. Factorial Trailing Zeroes

172. Factorial Trailing Zeroes

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Question

Total Accepted: 45801  Total Submissions: 147715  Difficulty: Easy

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

分析：

缩小问题规模找规律：1!=1,2!=2,3!=6,4!=24,5!=120,6!=720,7!=5040,8!=40320,9!=362880,10!=3628800
草，知道和5,10有关，但是没做对，看了下别人家的思路：
核心思想每一个尾巴上的0，绝对是2和5的结果
绝对规律1：2的个数绝对比5的个数多（这很显然）
绝对规律2：有多少个质因子5就有多少个尾巴0
26！有多少个质因子5？5，10，15，20，25（有的数贡献两个5，6个）
51！有多少个质因子5？5，10，15，20，25,30,35,40,45,50,（12个），最终规律：5的个数，n/5+n/25+n/125

class Solution {
public:
    int trailingZeroes(int n) {
        if(n<5)
            return 0;
        int cnt = 0;   //求质因子5的个数
           
        while(n/5 != 0) {    
            n /= 5;   
            cnt += n;   
        }   
           
        return cnt;   
    }
};

来源于别人家的具体分析：

Well, to compute the number of trailing zeros, we need to first think clear about what will generate a trailing 0? Obviously, a number multiplied by 10 will have a trailing 0 added to it. So we only need to find out how many 10's will appear in the expression of the factorial. Since10 = 2 * 5 and there are a bunch more 2's (each even number will contribute at least one 2), we only need to count the number of 5's.

Now let's see what numbers will contribute a 5. Well, simply the multiples of 5, like 5, 10, 15, 20, 25, 35, .... So is the result simply n / 5? Well, not that easy. Notice that some numbers may contribute more than one 5, like 25 = 5 * 5. Well, what numbers will contribute more than one 5? Ok, you may notice that only multiples of the power of 5 will contribute more than one5. For example, multiples of 25 will contribute at least two 5's.

Well, how to count them all? If you try some examples, you may finally get the result, which is n / 5 + n / 25 + n / 125 + .... The idea behind this expression is: all the multiples of 5 will contribute one 5, the multiples of 25 will contribute one more 5 and the multiples of 125 will contribute another one more 5... and so on. Now, we can write down the following code, which is pretty short.

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原文地址：http://blog.csdn.net/ebowtang/article/details/50443069

原作者博客：http://blog.csdn.net/ebowtang