public class Demo{ public static void main(String [] args){ int num=2; System.out.println(num++);//后置加加,先打印后加 2 System.out.println(num); System.out.println(++num);//前置加加,先加后打印 4 System.out.println(num); int num1=4; int result=num1++;//先赋值给result后num自
using System.Linq; using System.Text; using System.Threading.Tasks; namespace 第四天_加加减减 { class Program { static void Main(string[] args) { //int number = 10; //int result = 10+number++; //Console.WriteLine(number); //输出结果为11 //Console.WriteLine(resul
求阶乘序列前N项和 #include <stdio.h> double fact(int n); int main() { int i, n; double item, sum; while (scanf("%d", &n) != EOF) { sum = 0; if (n <= 12) { for (i = 1; i <= n; i++) { item = fact(i); sum = sum + item; } } printf("%.0f
来总结下求阶乘的各种方法哈. 写在最前:①各个代码仅仅是提供了求阶乘的思路,以便在实际须要时再来编码,代码并不健壮!②各个程序都在1到10内測试正确. 代码一: #include<iostream> using namespace std; int fac(int); int main() { int n; while(cin>>n) { cout<<n<<"!= "<<fac(n)<<endl; } return
'''Created on 2018年10月28日递归函数示例:阶乘'''def my_fun_example1(n): ''' 非递归函数求阶乘示例 ''' result = n for i in range(1,n): result *= i return resultdef my_fun_example2(n): ''' 递归函数求阶乘示例 ''' if n == 1: return 1 else
题目:求1+2!+3!+...+20!的和分析:使用递归求解 0的阶乘和1的阶乘都为1 public class Prog21{ public static void main(String[] args){ long sum=0L; for(int i=1;i<=20;i++) { sum+=factorial(i); } System.out.println(sum); } //递归求阶乘 public static long factorial(int n) { if(n==0||n==1
22 [程序 22 递归求阶乘] 题目:利用递归方法求 5!. 程序分析:递归公式:fn=fn_1*4! package cskaoyan; public class cskaoyan22 { @org.junit.Test public void recursion() { long number = 5; System.out.println(factorial(number)); } private long factorial(long number) { if (number == 1
21 [程序 21 求阶乘] 题目:求 1+2!+3!+...+20!的和 程序分析:此程序只是把累加变成了累乘. package cskaoyan; public class cskaoyan21 { @org.junit.Test public void sum() { int mix = 1; int max = 20; long total = 0; for (int i = mix; i <= max; i++) { total += factorial(i); } System.ou
