这题主要是去掉阶乘末尾的0。是个老题了，编程之美中就有讨论。因为0都是由2*5得来的。只要找出乘数中有多少个2*5对就行了。
因为2的个数远多于5，所以只要找出5的个数即可。因为n最大为4220,5的个数为:
n/5+n/5/5+n/5/5/5+n/5/5/5/5+n/5/5/5/5/5+n/5/5/5/5/5/5;
然后再去除相应数目的2。这样剩下的数只需两两相乘后取最后一位即可。
#include <iostream>
#include <fstream>
using namespace std;
ifstream in("fact4.in");
ofstream out("fact4.out");
void solve()
{
int n;
in>>n;
int numof5 = n/5+n/5/5+n/5/5/5+n/5/5/5/5+n/5/5/5/5/5+n/5/5/5/5/5/5;
int res = 1;
int tmp;
for(int i=1;i<=n;++i){
tmp = i;
while(tmp%5==0) tmp/=5;
while(tmp%2==0&&numof5!=0){
tmp/=2;
numof5--;
}
res*=tmp;
res%=10;
}
out<<res<<endl;
}
int main(int argc,char *argv[])
{
solve();
return 0;
}
原题：
Factorials
The factorial of an integer N, written N!, is the product of all
the integers from 1 through N inclusive. The factorial quickly becomes
very large: 13! is too large to store in a 32-bit integer on most
computers, and 70! is too large for most floating-point variables. Your
task is to find the rightmost non-zero digit of n!. For example, 5! =
1 * 2 * 3 * 4 * 5 = 120, so the rightmost non-zero digit of 5! is 2.
Likewise, 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040, so the rightmost non-zero
digit of 7! is 4.
PROGRAM NAME: fact4
INPUT FORMAT
A single positive integer N no larger than 4,220.
SAMPLE INPUT (file fact4.in)
7
OUTPUT FORMAT
A single line containing but a single digit: the right most non-zero
digit of N! .
SAMPLE OUTPUT (file fact4.out)
4