//求階乘最后非零位,復(fù)雜度O(nlogn)
//返回該位,n以字符串方式傳入
#include <string.h>
#define MAXN 10000

int lastdigit(char* buf){
	const 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;
}