这周Java课程有个小作业:Java递归实现从n个数中选取m个数的所有组合 代码如下: //其中 n 取 1,2,3,4,5 五个数, m 取 3 package javaText; public class text { static int N = 5; static int M = 3; static int[] a= new int[]{1,2,3,4,5}; static int[] b = new int[M]; public static void main(String[] ar
求集合里元素的个数 输出最大的个数是多少 Sample Input41 23 45 61 641 23 45 67 8 Sample Output42 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # include <queue> # define LL long long usi
day11 --------------------------------------------------------------- 实例018:复读机相加 题目 求s=a+aa+aaa+aaaa+aa-a的值,其中a是一个数字.例如2+22+222+2222+22222(此时共有5个数相加),几个数相加由键盘控制. 分析:很简单,字符串*x可以复制. 1 a = input('请输入数字:') 2 n = input("请输入要加几次:") 3 s = 0 4 for i in
返回本章节 返回作业目录 需求说明: 通过给定可变参数方法,计算任意给定的多个int类型数据之和. 实现思路: 定义可变形参方法,参数类型是int类型. 定义变量 sum 接受最终的和. 通过 for 循环计算最终的和值. 在主方法中调用可变参数方法,输出最终和值. 实现代码: public class TestCal { //定义求和方法 public int sum(int...nums) { int sum = 0; for (int num : nums) { sum += num; }
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 m
