循环从1乘到20,要注意的就是结果可能会很大,长度超出int类型的范围,所以定义乘积的时候用long. 代码如下: public class Practice3 { public static void main(String args[]){ long result=1; for(int i=1;i<=20;i++) result=result*i; System.out.println("20的阶乘为"+result); } } 效果如图:
package com.zf.s2;//创建一个包 import java.math.BigInteger;//导入类 import java.util.ArrayList; import java.util.List; public class TextFactorial {//操作计算阶乘的类 public static int simpleCircle(int num){//简单的循环计算的阶乘 int sum=1; if(num<0){//
chapter 20 注解 三种标准注解和四种元注解: 编写注解处理器 chapter 21 并发 基本的线程机制 定义任务 package cn.test; public class LiftOff implements Runnable { protected int countDown = 10; private static int taskCount=0; private final int id= taskCount++; public LiftOff() {} public Lif
import java.util.Scanner; public class n的阶乘 { public static void main(String[] args) { Scanner sc =new Scanner(System.in); int n = sc.nextInt(); int num[] = new int[10000000]; int i, j; if (n == 1 || n == 0) { System.out.println(1); } else { int p, h
Annotations They provide information that you need to fully describe your program, but that cannot be expressed in Java. In general the kind of annotations you add and what you do with them are entirely up to you. Java SE5 contains three generalpurpo
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
题目描述: Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)
