Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 53703 Accepted: 25237 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh
获取链表List中对象属性最大值最小值(Max,Min)的方法: 1.创建一个类,类中有一个属性A /// <summary> /// 用于测试属性的类 /// </summary> public class ListTest { private int a; public int A { get { return a; } set { a = value; } } } 2.在主函数中创建3个类A的对象,分别给属性A赋值为1,2,10,将3个对象加入链表中 class Progra
一.我个人觉得求最大值比较简单的一种方法(当然同时求最大值和最小值时稍微改改也能行) #include <stdio.h> int main(void) { int f, i, max; max = f; ; i<=; i++) //这里的i可以随便该 { scanf("%d", &f); //只定义一个f变量,用来接收输入的数字,减少的代码量 if(f>max) max = f; } printf("\nmax = %d\n", m
import java.util.Scanner; public class ArrayDemo { public static void main(String []args) { //------------------------------------------------------- //线性查找 int [] num ={10,20,30,40,50}; Scanner input1 = new Scanner(System.in); System.out.println("请输
1.求最大值 实例:小明买手机,询问了4家店的价格,分别是2800,2900,2600,3000,输出最低价 正确代码:类似打擂台,将数组第一个值赋给最小值,然后依次比较 import java.util.Scanner; public class Test1{ public static void main(String[] args){ Scanner input=new Scanner(System.in); System.out.println("询问了几家店,请输入"); i
