HUNAM 10493

 

Factorials Again

Time Limit: 2000ms, Special Time Limit:5000ms, Memory Limit:32768KB

Total submit users: 293, Accepted users: 243

Problem 10493 : No special judgement

Problem description

The factorial of an integer N, written N!, is the product of all the integers from 1 through N inclusive. The factorial quickly becomes very large: 13! is too large to store in a 32-bit integer on most computers, and 70! is too large for most floating-point variables. Your task is to find the rightmost non-zero digit of n!. For example, 5! = 1 * 2 * 3 * 4 * 5 = 120, so the rightmost non-zero digit of 5! is 2. Likewise, 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040, so the rightmost non-zero digit of 7! is 4.

Input

There are multiple test cases. Each contains A single positive integer N no larger than 10,000 in a single line. Input ends with a zero and this line should not be processed.

Output

A single line containing but a single digit: the right most non-zero digit of N! .

Sample Input

7 0

Sample Output

4     代码：注意进位问题，不能只取最后一位 #include <iostream> #include <cstdio> #include <cstring> using namespace std; int n; int main() { int sum; int i; while(scanf("%d",&n)!=EOF) { if(n==0) break; sum=1; for(i=2;i<=n;i++) { sum*=i; if(sum%10==0) { while(sum%10==0) sum/=10; } if(sum>=10000) { sum%=10000; } } while(sum%10==0) { sum/=10; } printf("%d\n",sum%10); } return 0; }