package test; import java.math.BigInteger; import java.util.ArrayList; import java.util.List; import java.util.Scanner; public class MyTest { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); List<Int…
import java.util.Scanner; public class Factorial { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n; while (true) { System.out.print("请输入一个正整数(输入0退出循环):"); try { n = sc.nextInt(); } catch (Exception e) { System…
题目 Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 分析 Note中提示让用对数的时间复杂度求解,那么如果粗暴的算出N的阶乘然后看末尾0的个数是不可能的. 所以仔细分析,N! = 1 * 2 * 3 * ... * N 而末尾0的个数只与这些乘数中5和2的个数有关,因为每出现一对5和2就会产生…
public class Factorial{ public static void main(String[] args){ for (int i = -5; i <= 5; i++) { System.out.printf("%d! = %d\n", i, negative(i)); } } // 判断n是否为负数 public static int negative(int n){ if (n >= 0){ return factorial(n); } else {…
在JAVA中求阶乘首先遇到的问题就是结果溢出,不管是使用int还是long,double都无法表示1000!这么大的天文数字,这里暂且用BigInteger解决这个问题! 下面是使用递归和尾递归分别计算1000的阶乘: import java.math.BigInteger; public class Main { public static void main(String[] args) { long t = System.currentTimeMillis(); System.out.pr…
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 53785 Accepted Submission(s): 15217 Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N! Input One N in one…
我记得有一份代码是非常有技巧的,然而这一份就是很死板-每次跑50000,因为10000的阶乘最多才50000位,这样肯定就过了 #include<cstdio> #include<string.h> #include<iostream> #include<algorithm> using namespace std; const int maxn=50000; int f[maxn+1]; int main() { int n,c,k; while(~sca…
转载请注明出处:http://blog.csdn.net/lttree Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4436 Accepted Submission(s): 2642 Problem Description Now our hero finds the…
转载请注明出处:http://blog.csdn.net/lttree Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4436 Accepted Submission(s): 2642 Problem Description Now our hero finds the…
1.设置递归层数 #设置recursion函数的层数,默认是100层 import sys sys.setrecursionlimit(10000) 2. 阶乘 #定义一个阶乘函数 def factorial(n): result = n for i in range(1,n): result *= i return result number = int(input('请输入一个正整数:')) result = factorial(number) print("%d的阶乘是:%d"…
