1024！末尾有几个零？解决方法

1024！末尾有几个零？
1024！末尾有几个零？

[解决办法]

C/C++ code

#include <stdio.h>int totalzero(int n){    int total = 0;    while (n > 5)    {        n = (n - (n % 5)) / 5;        total += n;    }    return total;}void main(){    printf("%d\n", totalzero(1024));}
[解决办法]
两个办法：
1. 暴力做法：用大数计算1024！，然后再看后面有多少个0。大数计算的代码可以参考：
大数的四则运算及求模 

2. 巧慧做法：看1到1024这些数中，有多少个个位是0和5的数字，比如
10！ = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10
因为5和其前面的任何一个偶数相乘都会产生0，个位本身是0的情况就更不用说了。所以10！最后有2个0.

由此可以推知15！最后有3个0...

当然特殊的情况需要注意，比如数字是100,1000这样的情况，还有50,500这样的情况，楼主稍加思考不难得出正确的结论。
[解决办法]
Python code
m=input("Calculate N!, input N:")n, r = 1, 1while n <= m:    r=r*n    n=n+1print r
[解决办法]
http://book.51cto.com/art/200909/150056.htm
上面的链接很适合楼主……
或者，楼主google一下“不要被阶乘吓到”
[解决办法]
赵哥v5……
公式：1024/5+1024/25+1024/125+1024/625
=204 + 40 + 8 + 1 =253
[解决办法]
记得小学数学奥林匹克题目是这样的，
答案确实是253
算法简单：10 = 2 *5，必须有5的因子才可能乘到10；
所以计算1到1024中分解出5因子个数即得到结果。
[解决办法]
There is a formula for finding the number of zeros a factorial number will have.  
It is:  

     floor(n/5) + floor(n/25) + floor(n/125) + ... + floor(n/5^n) + ...  

The floor function means round down to the nearest integer.  

For example, let's take 5!, or n = 5.  

     floor(5/5) = floor(1) = 1.  

So, 5! ends in 1 zero.  

Let's go to 15!  

     floor(15/5) = floor(3) = 3  

So, 15! ends in 3 zeros.  

Let's move on to 100!.  

     floor(100/5) + floor(100/25) = floor(20) + floor(4) = 20 + 4 = 24  

If you went to 200, you would have to use the next power of 5, or 5^3 = 125.  




The reason this works is that a number ends in 0 if it is divisible by 10. Divisible by 10 means divisible by both 5 and 2. But there are lots of numbers divisible by 2 (half of them). So, we concentrate on the number of times a number is divisible by 5. But there are tricky numbers like 25, which are divisible by 5 twice, so we have to take those into account (or floor(n/25)). Then again, there are numbers that are divisible by 5 three times, like 125, so we have to take them into account.