Last non-zero Digit in N!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6276 Accepted Submission(s): 1556
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
求N!的最后一个非零数字，套模板。
#include
#include
#define MAXN 10000 int a[MAXN]; char str[MAXN]; 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 lastdigit(char *buf){ int len = strlen(buf), 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", str) == 1){ printf("%d\n", lastdigit(str)); } return 0; }