UVA 568 (13.07.28)

来源：互联网发布：淘宝网中华活胰宝编辑：程序博客网时间：2024/04/27 19:59

Just the Facts

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

NN!011122364245120103628800

For this problem, you are to write a program that can compute the lastnon-zero digit of anyfactorial for ( $0 \le N \le 10000$ ). For example, if your program is asked tocompute the last nonzerodigit 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 not exceeding 10000,each on its own linewith no other letters, digits or spaces. For each integerN, you shouldread the value and compute the last nonzero digit of N!.

Output

For each integer input, the program should print exactly one line ofoutput. Each line of outputshould contain the valueN, right-justified in columns 1 through 5 withleading blanks, not leadingzeroes. Columns 6 - 9 must contain `` ->" (space hyphen greater space).Column 10 must contain the single last non-zero digit ofN!.

Sample Input

122612531259999

Sample Output

    1 -> 1    2 -> 2   26 -> 4  125 -> 8 3125 -> 2 9999 -> 8

题意:

给出N, 求N的阶乘的最后一位是什么, 除零以外

思路:

考虑到零都是由2*5得来的, 那么每一次阶乘的时候, 把要乘的数对2和5去判断是否整除, 并统计整除2和5的次数分别是多少

最后我们再看看对2和5整除的次数分别是多少, 哪个多就还原哪个, 还原到对2对5的整除次数一致为止, 保证了2*5=10的成对性

AC代码如下:

#include<stdio.h>int main() {int N;int ans, t, num2, num5;while(scanf("%d", &N) != EOF) {ans = 1;num2 = num5 = 0;for(int i = N; i > 1; i--) {t = i;while(t % 2 == 0) {num2++;t = t / 2;}while(t % 5 == 0) {num5++;t = t / 5;}ans = ans * t;ans = ans % 10;}if(num2 > num5) {while(num2 > num5) {ans = ans * 2;ans = ans % 10;num2--;}}if(num5 > num2) {while(num5 > num2) {ans = ans * 5;ans = ans % 10;num5--;}}printf("%5d -> %d\n", N, ans);}return 0;}