| Just the Facts |
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 lastnon-zero digit of anyfactorial for (
). 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 integer N, 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 value N, 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 of N!.
Sample Input
1
2
26
125
3125
9999
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;
}