常熟java在哪培訓(xùn)_常熟JAVA培訓(xùn)
來源：教育聯(lián)展網(wǎng) 編輯：佚名 發(fā)布時間：2018-08-21
Java工程師就業(yè)前景
Java工程師就業(yè)前景
2015年，在美國、加拿大、澳大利亞、新加坡等發(fā)達(dá)國家和中等發(fā)達(dá)國家，
JAVA軟件工程師年薪均在4—15萬美金，而在國內(nèi)，JAVA軟件工程師也有極好的工作機(jī)會和很高的薪水。
在未來5年內(nèi)，合格軟件人才的需求將遠(yuǎn)大于供給。JAVA軟件工程師是目前
國際高端計算機(jī)領(lǐng)域就業(yè)薪資非常高的一類軟件工程師。
一般情況下的JAVA軟件工程師是分四個等級，從軟件技術(shù)員到助理軟件工程
師，再到軟件工程師，**后成為高級軟件工程師。
根據(jù)IDC的統(tǒng)計數(shù)字，在所有軟件開發(fā)類人才的需求中，對JAVA工程師的需
求達(dá)到全部需求量的60%—70%。同時，JAVA軟件工程師的工資待遇相對較高。
通常來說，具有3—5年開發(fā)經(jīng)驗的工程師，擁有年薪15萬元是很正常的一個
薪酬水平。80%的學(xué)生畢業(yè)后年薪都超過了8萬元。
根據(jù)專業(yè)數(shù)據(jù)分析，由于我國經(jīng)濟(jì)發(fā)展不均衡因素，JAVA軟件工程師工資待
遇在城市之間的差異也較大，一級城市(如北京、上海等)，初級軟件工程師的待遇大概在4000-6000之間，中級軟件工程師的待遇在6000—8000之間，
而高級軟件工程師的待遇基本破萬。
JavaWeb開發(fā)
-
01HTML5與CSS3
-
1．B/S架構(gòu)
-
2．HTML基本使用
-
3．HTML DOM
-
4．CSS選擇器
-
5．常用樣式
-
6．盒子模型與布局
-
7．HTML5新特性
-
8．CSS3新特性
-
02JavaScript
-
03jQuery
-
1．jQuery快速入門
-
2．jQuery語法詳解
-
3．jQuery核心函數(shù)
-
4．jQuery對象/JavaScript對象
-
5．jQuery選擇器
-
6．jQuery 文檔處理
-
7．jQuery事件
-
8．jQuery動畫效果
-
04AJAX&JSON
-
05XML
-
1．XML用途
-
2．XML文檔結(jié)構(gòu)
-
3．XML基本語法
-
4．DOM&SAX解析體系
-
5．DOM4j節(jié)點查詢
-
6．DOM4j文檔操作
-
7．xPath語法
-
8．xPath快速查詢
-
06bootstrap
-
1．bootstrap快速使用
-
2．柵格系統(tǒng)
-
3．表單、表格、按鈕、圖片
-
4．下拉菜單
-
5．按鈕組使用
-
6．導(dǎo)航條
-
7．分頁、進(jìn)度條
-
07Web服務(wù)器基礎(chǔ)
-
1．HTTP協(xié)議
-
2．HttpWatch
-
3．Tomcat服務(wù)器搭建
-
4．Tomcat目錄結(jié)構(gòu)解析
-
5．Tomcat端口配置
-
6．Tomcat啟動&停止
-
7．Tomcat&Eclipse整合
-
8．Eclipse配置優(yōu)化
-
08Servlet
-
09JSP
-
1．JSP語法
-
2．JSP原理
-
3．JSP腳本片段&表達(dá)式
-
4．JSP聲明&指令
-
5．JSP九大隱含對象
-
6．域?qū)ο笫褂?/span>
-
10JSTL
-
1．JSTL簡介
-
2．JSTL-核心標(biāo)簽庫
-
3．JSTL-函數(shù)標(biāo)簽庫
-
4．JSTL-fmt標(biāo)簽庫
-
5．自定義標(biāo)簽庫使用
-
6．自定義標(biāo)簽庫原理
-
11EL
-
1．EL表達(dá)式簡介
-
2．EL使用
-
3．EL取值原理
-
4．EL的11大隱含對象
-
5．EL2.2與3.0規(guī)范
-
6．EL邏輯運(yùn)算
-
7．函數(shù)庫深入
-
12Cookie&Session
-
1．Cookie機(jī)制
-
2．Cookie創(chuàng)建&使用
-
3．Session原理
-
4．Session失效
-
5．Url重寫
-
6．Session活化&鈍化
-
7．Token令牌應(yīng)用
-
13Filter&Listener
-
1．Filter原理
-
2．Filter聲明周期
-
3．Filter鏈
-
4．Filter登錄驗證
-
5．Filter事務(wù)控制
-
6．Listener原理
-
7．八大監(jiān)聽器使用
-
8．Listener監(jiān)聽在線用戶
-
14國際化
-
1．國際化原理
-
2．ResourceBundle&Locale
-
3．國際化資源文件
-
4．日期/數(shù)字/貨幣國際化
-
5．頁面動態(tài)中英文切換
-
6．頁面點擊鏈接中英文切換
-
7．fmt標(biāo)簽庫的使用
-
15文件上傳
HDU 6016
>
Count the Sheep
Time Limit: 3000/1500 MS (java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 852 Accepted Submission(s): 362 PRoblem Description
Altough Skipping the class is happy, the new term still can drive luras anxious which is of course because of the tests! Luras became worried as she wanted to skip the class, as well as to attend the BestCoder and also to prepare
for tests at the same time.
However, As the result of preparing for tests, luras had no time to practice programing. She didn t want to lose her rating after attending BC. In the end, she found BCround92 s writer snowy_smile for help, asking him to leak her something.
Snowy_smile wanted to help while not leaking the problems. He told luras, the best thing to do is to take a good rest according to the following instructions first.
"Imagine you are on the endless grassland where there are a group of sheep. And n sheep of them are silent boy-sheep while m sheep are crying girl-sheep. And there are k friend-relationships between the boy-sheep and girl-sheep.Now You can start from any sheep,
keep counting along the friend relationship. If you can count 4 different sheep, you will exceed 99% sheep-counters and fall asleep."
Hearing of the strange instructions, luras got very shocked. Still, she kept counting. Sure enough, she fell asleep after counting 4 different sheep immediately. And, she overslept and missed the BestCoder in the next day. At a result, she made it that not
losing her rating in the BCround92!!!
However, you don t have the same good luck as her. Since you have seen the 2nd problem, you are possible to have submitted the 1st problem and you can t go back.
So, you have got into an awkward position. If you don t AC this problem, your rating might fall down.
You question is here, please, can you tell that how many different 4-sheep-counting way luras might have before her sleep?
In another Word, you need to print the number of the "A-B-C-D" sequence, where A-B, B-C, C-D are friends and A,B,C,D are different.
Input
The first line is an integer T which indicates the case number.
and as for each case, there are 3 integers in the first line which indicate boy-sheep-number, girl-sheep-number and friend-realationship-number respectively.
Then there are k lines with 2 integers x and y in each line, which means the x-th boy-sheep and the y-th girl-sheep are friends.
It is guaranteed that——
There will not be multiple same relationships.
1 <= T <= 1000
for 30% cases, 1 <= n, m, k <= 100
for 99% cases, 1 <= n, m, k <= 1000
for 100% cases, 1 <= n, m, k <= 100000
Output
As for each case, you need to output a single line.
there should be 1 integer in the line which represents the number of the counting way of 4-sheep-sequence before luras s sleep.
Sample Input
3
2 2 4
1 1
1 2
2 1
2 2
3 1 3
1 1
2 1
3 1
3 3 3
1 1
2 1
2 2
Sample Output
8
0
2
已經(jīng)AC過的代碼：
#include<cstdio>
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
long long d1[100010],d2[100010];
struct edge
{
int x,y;
};
edge e[100010];
int main()
{
int t,n,m,k;
scanf("%d",&t);
while(t--)
{
long long ans=0;
memset(d1,0,sizeof(d1));
memset(d2,0,sizeof(d2));
scanf("%d %d %d",&n,&m,&k);
for(int i=0;i<k;i )
{
scanf("%d %d",&e[i].x,&e[i].y);
d1[e[i].x] ;
d2[e[i].y] ;
}
for(int i=0;i<k;i )
{
ans =(d1[e[i].x]-1)*(d2[e[i].y]-1);
}
cout<<ans*2<<endl;
}
return 0;
}
相關(guān)推薦：
南京java培訓(xùn) 南京java培訓(xùn)班 南京java培訓(xùn)機(jī)構(gòu)
上一篇：張家港java編程需要學(xué)什么_張家港JAVA培訓(xùn) 下一篇：昆山零基礎(chǔ)java開發(fā)培訓(xùn)_昆山JAVA培訓(xùn)
