HDOJ 1066-Last non-zero Digit in N!【阶乘最后非零解：模板】

Last non-zero Digit in N!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7146    Accepted Submission(s): 1797

Problem Description

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
   
   
   
   

 

Source

South Central USA 1997

 

解题思路：

就是求阶乘的最后非零解，总结摸板，以后可以直接用。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 10000
using namespace std;
char map[100000];
int lastdigit(char *buf)
{
	int mod[20]={1,1,2,6,4,2,2,4,2,8,4,4,8,4,6,8,8,6,8,2};
	int len=strlen(buf),a[maxn],i,c,ret=1;
	if(len==1)
	{
		return mod[buf[0]-'0'];
	}
	for(i=0;i<len;i++)
	{
		a[i]=buf[len-1-i]-'0';
	}
	for(;len;len-=!a[len-1])
	{
		ret=ret*mod[a[1]%2*10+a[0]]%5;
		for(c=0,i=len-1;i>=0;i--)
		{
			c=c*10+a[i],a[i]=c/5,c%=5;
		}
	}
	return ret+ret%2*5;
}
int main()
{
	while(scanf("%s",map)!=EOF)
	{
		printf("%d\n",lastdigit(map));
	}
	return 0;
}