委托的定义: (1) 将方法作为变量使用的一种机制,就是将方法当作变量用(声明,赋值,传参) (2) 将变量当作方法来用,首先就要去声明变量,就要考虑变量的类型,就是(委托变量,对应方法的返回值,参数等),顾名思义:委托就是委托别人去干某些事情,下面是一个实例方法的实现 例如:我现在饿了,但是我很懒,就是不想出去买,所以这时候我可以委托室友帮我带一份,这样我就已经实现委托了 class Program { //声明一个委托 delegate int DelegateCompare(objec
Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:Given m satisfies the following const
# include <stdio.h> # define N main(){ int a, b; ,,,,,,,,,,,,,,,,}; //array中输入需要排序的数字 ]; ; a < N; a++){ if(array[a]>max){ max = array[a]; //使用max函数,快速筛选出最大值 } } printf("Max = %d\n",max,b); }
Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:If n is the length of array, assume t
Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:Given m satisfies the following const
总结:输出最大值,最小值,可以先排序,再输出就容易点了.不用循环判断.我还没学.但是觉得很好用 package com.aini; import java.util.Arrays; import java.util.Scanner; //求把数组里的元素,输出最大值,最小值,平均值 public class PAI { public static void main(String[] args) { Scanner c = new Scanner(System.in); System.out.p
给定一个非负整数数组和一个整数 m,你需要将这个数组分成 m 个非空的连续子数组.设计一个算法使得这 m 个子数组各自和的最大值最小.注意:数组长度 n 满足以下条件: 1 ≤ n ≤ 1000 1 ≤ m ≤ min(50, n)示例:输入:nums = [7,2,5,10,8]m = 2输出:18解释:一共有四种方法将nums分割为2个子数组.其中最好的方式是将其分为[7,2,5] 和 [10,8],因为此时这两个子数组各自的和的最大值为18,在所有情况中最小.详见:https:
