集成运放小解1,我们首先要去理解虚短、虚断。这是构成运放的始终①虚断:2、3两点之间断路,无电流(多少有点)虚短:2、3两点电压一样,无压差(多少有点)②为什么运
https://www.u72.net/daima/nhrua.html - 2024-08-02 21:05:25 - 代码库集成运放小解(一)我们首先要去理解虚短、虚断。这是构成运放的始终①虚断:2、3两点之间断路,无电流(多少有点)虚短:2、3两点电压一样,无压差(多少有点)②为什么运
https://www.u72.net/daima/nhr27.html - 2024-08-02 21:19:14 - 代码库题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1229解题思路,最开始想的是把输入的数据存入数组中,比如输入 1,在数组中就储存为1000(因为数据不超
https://www.u72.net/daima/nn13h.html - 2024-07-31 22:22:46 - 代码库相遇周期Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3382 Accepted Submissio
https://www.u72.net/daima/nvs5r.html - 2024-10-30 06:24:02 - 代码库/*A + B Problem IIProblem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Su
https://www.u72.net/daima/nurnh.html - 2024-10-22 23:47:39 - 代码库框架和代码都是死的,人是活的。有一个靠谱的初创技术团队和核心技术大拿,知道什么阶段用什么技术方案来满足业务和可扩展性之间的平衡,很重要。比如说,
https://www.u72.net/daima/nus9m.html - 2024-10-23 11:15:02 - 代码库/*a/b &#43; c/dProblem Description给你2个分数,求他们的和。并要求和为最简形式。 Input输入首先包括一个正整数T(T<=1000),表示有T组測试数据,然后是T
https://www.u72.net/daima/ns35b.html - 2024-10-18 14:31:39 - 代码库变形课Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 18343 Accepted Submission
https://www.u72.net/daima/nu41z.html - 2024-10-25 15:57:39 - 代码库01背包问题:有一个体积为V的背包,有n件物品,每件物品的体积,价&#20540;分别为w[i],p[i];要从n件物品中选些放入背包中,使背包里物品的总价&#20540;最大。动
https://www.u72.net/daima/nsm62.html - 2024-10-20 15:32:39 - 代码库题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2000注意过滤掉输入三个字符之后的回车键即可#include<stdio.h>int main(){ char a,b,c,t; while
https://www.u72.net/daima/nnku5.html - 2024-07-31 10:52:23 - 代码库这一题目是要求连续子序列的最大和,所以在看到题目的一瞬间就想到的是把所有情况列举出来,再两个两个的比较,取最大的(即为更新最大值的意思),这样的思路很
https://www.u72.net/daima/na115.html - 2024-07-30 20:28:34 - 代码库Number SequenceTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11925 Accepted S
https://www.u72.net/daima/nbz1a.html - 2024-08-05 19:12:58 - 代码库解题思路:先忽略饭卡余额大于等于5块才能买饭这一细节,需要求的是饭卡里面剩余的钱最少,转化一下,变成花的钱最多,那么剩下的钱就最少,再考虑余额大于等于5块
https://www.u72.net/daima/nbz4f.html - 2024-08-05 19:18:48 - 代码库骨牌铺方&#26684;Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30887 Accepted S
https://www.u72.net/daima/nbz71.html - 2024-08-05 19:25:13 - 代码库继续畅通projectTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17814 Accept
https://www.u72.net/daima/nba71.html - 2024-10-02 07:29:01 - 代码库小明A+BTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 43915 Accepted Submissio
https://www.u72.net/daima/nbbsn.html - 2024-10-03 00:47:01 - 代码库分拆素数和Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39476 Accepted Submis
https://www.u72.net/daima/nb4kb.html - 2024-10-04 18:08:02 - 代码库无线电波波段划分波段名称波长范围(m)频段名称频率范围超长波长波中波短波1,000,000~10,00010,000~1,0001,000~100100~~1010~11~0.10.1~0.010.01~0.00
https://www.u72.net/daima/nwh3h.html - 2024-11-04 07:29:39 - 代码库文/爱跑步的光头哥其他传统企业的机会我都没考虑,来万达<em>电</em>商是因为这次确实不一样——大时代面前一个很特别的机会。全球最大的O2O试验有几家比较大的传
https://www.u72.net/daima/9def.html - 2024-07-27 06:25:20 - 代码库一步步打造一个简单的 MVC <em>电</em>商网站 - BooksStore(二) 本系列的 GitHub地址:https://github.com/liqingwen2015
https://www.u72.net/daima/8f6h.html - 2024-09-11 13:16:13 - 代码库