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

这道题，嘿嘿，套用高精模板，每次计算后把后面的0擦去，然后再对10000取余即可……

因为长度问题，我就不贴我的高精模板了……

/*
USER: wangyuc2
TASK: fact4
LANG: C++

Compiling...
Compile: OK

Executing...
    Test 1: TEST OK [0.022 secs, 2868 KB]
    Test 2: TEST OK [0.000 secs, 2872 KB]
    Test 3: TEST OK [0.000 secs, 2868 KB]
    Test 4: TEST OK [0.011 secs, 2872 KB]
    Test 5: TEST OK [0.011 secs, 2872 KB]
    Test 6: TEST OK [0.054 secs, 2872 KB]
    Test 7: TEST OK [0.076 secs, 2868 KB]
    Test 8: TEST OK [0.065 secs, 2868 KB]
    Test 9: TEST OK [0.076 secs, 2872 KB]
    Test 10: TEST OK [0.076 secs, 2872 KB]

All tests OK.
Your program ('fact4') produced all correct answers!  This is your
submission #2 for this problem.  Congratulations!

*/
#include <fstream>
#include <iostream>
#include <cstring>
#include <memory>
#include <algorithm>
#define cin fin
using namespace std;

string operator+(string x, string y); 

string operator-(string x, string y); 
string operator*(string x, string y); 
string operator*(string s, int a); 

string operator/(string s, int a); 
int     operator%(string s, int a); 
string operator/(string x, string y); 
string operator%(string x, string y); 

ifstream fin("fact4.in");
ofstream fout("fact4.out");

int main()
{
     string s,s1;
     int i,j,k,n,m,maxnum=0;
     cin>>n;
     s.assign("1");
     s1.assign("10000");
     for(i=1;i<=n;i++)
     {
         s=s*i;
         j=s.size()-1;
     //     cout<<s<<' ';
         while(s[j]=='0') {s.erase(j,j+1);j--;}
    //      cout<<s<<endl;
         s=s%s1;
     }
     fout<<s[s.size()-1]<<endl;
// system("PAUSE");
     return 0;
}