USACO Section 3.2 Factorials (fact4)

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

思路：hdu上有比它更变态的题，给个网址，上面讲的很详细。
http://blog.csdn.net/yibcs/article/details/8040862

Executing...
   Test 1: TEST OK [0.000 secs, 3228 KB]
   Test 2: TEST OK [0.000 secs, 3228 KB]
   Test 3: TEST OK [0.000 secs, 3228 KB]
   Test 4: TEST OK [0.000 secs, 3228 KB]
   Test 5: TEST OK [0.000 secs, 3228 KB]
   Test 6: TEST OK [0.000 secs, 3228 KB]
   Test 7: TEST OK [0.000 secs, 3228 KB]
   Test 8: TEST OK [0.000 secs, 3228 KB]
   Test 9: TEST OK [0.000 secs, 3228 KB]
   Test 10: TEST OK [0.000 secs, 3228 KB]

All tests OK.

/*
ID:wuhuaju2
PROG:fact4
LANG:C++
*/

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;

const int x[]={6,6,2,6,4,2,2,4,2,8,4,4,8,4,6,8,8,6,8,2};
const int maxn=100;
int a[maxn+10];
char s[maxn];
int n,t,l,e,beg,ans,tt;



void close()
{
    fclose(stdin);
    fclose(stdout);
    exit(0);
}

void work()
{
}

void init ()
{
freopen("fact4.in","r",stdin);
freopen("fact4.out","w",stdout);
    scanf("%s",s);
     l=strlen(s);
     if (l==1 && s[0]=='1')
     {
         cout<<1<<endl;
         close();
     }
     ans=1; int cnt=0;
     e=maxn; beg=e-l+1;
    // printf("e:%d beg:%d \n",e,beg);
     for (int i=l-1;i>=0;i--)
     {
         a[e-cnt]=s[i]-'0';
         cnt++;
     }
     while (beg<=e)
     {
         t=a[(e-1)] % 2 * 10+a[e];
         ans=(ans*x[t]) % 10;
         t=0;
         for (int i=e;i>=beg;i--)
         {
             t=a[i]*2 /10;
             a[i]=(a[i]*2+tt) % 10;
             tt=t;
         }
         if (t>=1)
         {
             beg--;
             a[beg]=1;
         }
         e--;
     }
     cout<<ans<<endl;
}

int main ()
{
    init();
    work();
    close();
    return 0;
}