ACM解题报告--Just the Facts (浙大1222)

Just the Facts

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Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 1052   Accepted Submit: 179  

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The expression N!, read as "N factorial," denotes the product of the first N positive integers, where N is nonnegative. So, for example,

N N!
0 1
1 1
2 2
3 6
4 24
5 120
10 3628800

For this problem, you are to write a program that can compute the last non-zero digit of the factorial for N. For example, if your program is asked to compute the last nonzero digit of 5!, your program should produce "2" because 5! = 120, and 2 is the last nonzero digit of 120.

Input

Input to the program is a series of nonnegative integers, each on its own line with no other letters, digits or spaces. For each integer N, you should read the value and compute the last nonzero digit of N!.

Output

For each integer input, the program should print exactly one line of output containing the single last non-zero digit of N!.

Sample Input

1
2
26
125
3125
9999

Sample Output

1
2
4
8
2
8

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Problem Source: South Central USA 1997  

该题与北大的1604类似，但是数据n可以达到10^300,用数组实现显然是不现实的,必须采用高精度算法。

代码如下:

#include <iostream>
#include <string>
using namespace std;

int data[5]={1,2,4,8,6};

int op(int b[],int k)
{
 int i,l=0;
 for(i=1;i<=k;i++)
  l=(b[i]+l*10)%4;
 if(!l)
  l=4;
 return data[l];
}

int compute(int a[],int n)
{
 int b[310],i,flag,l,k,x,t;
 flag=l=k=0;
 for(i=1;i<=n;i++)
 {
  x=a[i]+l*10;
  if(x/5)
   flag=1;
  if(flag)
   b[++k]=x/5;
  l=x%5;
 }
 x=1;
 if(!k)
 {
  for(i=1;i<=l;i++)
   x=x*i%10;
 }
 else
 {
  x=op(b,k)*compute(b,k)%10;
  t=(a[n]>=5?5:0);
  for(i=1;i<=l;i++)
   x=x*(i+t)%10;
 }
 return x;
}

int main()
{
 string s;
 while(cin>>s)
 {
  int a[310],i,n;
  n=s.length();
  for(i=0;i<n;i++)
   a[i+1]=s[i]-48;
  cout<<compute(a,n)<<endl;
 }
 return 0;
}