UVA - 568 - Just the Facts (简单数论!)
2014-12-07 21:44
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UVA - 568 Just the Facts
Description
). 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. InputInput 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!.OutputFor 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 Input1 2 26 125 3125 9999 Sample Output1 -> 1 2 -> 2 26 -> 4 125 -> 8 3125 -> 2 9999 -> 8 Miguel A. Revilla 1998-03-10 Source Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Number Theory :: Factorial Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Number Theory ::Factorial Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 5. Mathematics :: Factorial Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 1. Elementary Problem Solving :: Maths - Number Theory |
#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <cstdlib> #include <queue> #include <stack> #include <set> #include <algorithm> #define LL long long using namespace std; int main() { int n; while(scanf("%d", &n)!=EOF) { int ans=1; for(int i=1; i<=n; i++) { ans*=i; while(ans%10==0) { ans/=10; } ans%=100000; } printf("%5d -> %d\n", n, ans%10); } return 0; }
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