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| 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 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 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. Each line of output should contain the value N, 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 of N!.
Sample Input
122612531259999
Sample Output
1 -> 1
2 -> 2 26 -> 4 125 -> 8 3125 -> 2 9999 -> 8
Miguel A. Revilla
1998-03-10
1 /************************************************************************* 2
> File Name: 12345.cpp 3
> Author: acmicpcstar 4
> Mail: acmicpcstar@gmail.com 5
> Created Time: 2014年04月24日 星期四 11时46分18秒 6 ************************************************************************/ 7 8 #include9 #include 10 #include 11 #include 12 #include 13 using namespace std;14 const double pi=atan(1.0)*4.0;15 int main()16 {long long n,sum,i;17 while(cin>>n)18 {sum=1;19 for(i=2;i<=n;i++)20 {sum=sum*i;21 while(sum%10==0) sum/=10;22 sum=sum%1000000000;23 }24 printf("%5d -> %d\n",n,sum%10);25 }26 return 0;27 }
开始就保留了1位数字%10。。。结果发现会数据丢失。。。3125。。。后来COld。H说保留10位。。。。为了不溢出用了long long。。。网上说。。。5位就够了～哈哈～
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