先说需求:找出一个对象List中,某个属性值最大的对象. 1.定义对象 private class A { public int ID { get; set; } public string Name { get; set; } } 2.为两种方法定义两个时间段全局变量. 1 private static TimeSpan compare = new TimeSpan(); private static TimeSpan order = new TimeSpan(); 3.第一种方法:对列表
题目描述: In a given integer array nums, there is always exactly one largest element. Find whether the largest element in the array is at least twice as much as every other number in the array. If it is, return the index of the largest element, otherwise
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace _08求数组的最大值 { public delegate int DelCompare(object o1, object o2); class Program { static void Main(string[] args) { , , , ,
// 不用大与小与号,求两数最大值 #include <stdio.h> int max(int a, int b) { int c = a - b; int d = 1 << 31; if ((c&d) == 0) { return a; } else { return b; } } int main() { printf("%d是大数\n", max(0, 2)); printf("%d是大数\n", max(3, 4)); pr
Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 237978 Accepted Submission(s): 56166 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su
比如{3,2,4,3,6} 可以分成 {3,2,4,3,6} m=1; {3,6}{2,4,3} m=2 {3,3}{2,4}{6} m=3 所以m的最大值为3. bool isShare(int* a, int* group, int len, int m, int groupSize, int groupId, int curSize) { if (curSize == 0) { groupId++; curSize = groupSize; if (groupId == m + 1) {
总结:完全搞不懂,行和列是怎么弄的,,,,, package com.c2; import java.util.Scanner; public class Oaa { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n; System.out.print("请输入方阵的行与列:"); n = in.nextInt(); int[][] a = new int[n][n];
Hawk-and-Chicken Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2409 Accepted Submission(s): 712 Problem Description Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But t
此题用深搜很快就解决,我用宽度搜索和优先队列仅仅是为了练习他们的用法:深搜法在注释内: 1 #include<iostream> #include<cstring> #include<queue> using namespace std; ; int map[N][N]; int vis[N][N]; int n; struct Node { int x,y,value; friend bool operator<(Node a,Node b) {return a
1036: [ZJOI2008]树的统计Count Time Limit: 10 Sec Memory Limit: 162 MB Submit: 8421 Solved: 3439 [Submit][Status][Discuss] Description 一棵树上有n个节点,编号分别为1到n,每一个节点都有一个权值w.我们将以以下的形式来要求你对这棵树完毕一些操作: I. CHANGE u t : 把结点u的权值改为t II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值