Just the Facts
The expression N!, read as ``N factorial," denotes the product of the firstN positive integers, whereN 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 any factorial for (). 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 not exceeding 10000, each on its own line with no other letters, digits or spaces. For each integerN, 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. Each line of output should contain the valueN, right-justified in columns 1 through 5 with leading blanks, not leading zeroes. Columns 6 - 9 must contain `` ->" (space hyphen greater space). Column 10 must contain the single last non-zero digit ofN!.Sample Input
1 2 26 125 3125 9999
Sample Output
1 -> 1 2 -> 2 26 -> 4 125 -> 8 3125 -> 2 9999 -> 8
Miguel A. Revilla
1998-03-10
#include <stdio.h>
#include <string.h>
int main()
{
int i,j,n,m,s,t;
while(scanf("%d",&n)!=EOF)
{
m=1;
for(i=1;i<=n;i++)
{
m=m*i;
while(1)
{
t=m%10;
if(t==0)
{
m=m/10;
}else
{
break;
}
}
m=m%100000;
}
m=m%10;
printf("%5d -> %d\n",n,m);
}
return 0;
}