斯坦福java課程實(shí)驗(yàn)報(bào)告2

1、“java語(yǔ)言與internet程序設(shè)計(jì)”實(shí)驗(yàn)報(bào)告報(bào)告書姓名俞加平指導(dǎo)教師王肖虹學(xué)號(hào)1463313451丨期2016/4/5實(shí)驗(yàn)名稱assignment2實(shí)驗(yàn)內(nèi)容即題目?jī)?nèi)容problem 1problem 2：problem 3：problem 4：problem 5：problem 6：pyramidtargetprogramhierarchypythagoreantheoremfindrangehailstone數(shù)據(jù)代碼即運(yùn)行結(jié)果截圖problem 1:程序代碼/* file: pyramidtest.java* assignment: assignment 2* description

2、: |hi|_個(gè)金字塔* author: yujiaping* date: 2016-4-03*/importmcm grmphicg *;gram.*;publicclasspyramidtestextends graphicsprogram publicvoid run ()/設(shè)置窗口的大小setsize (win_widthf win_height);/從下到上依次畫出每行方塊for(inti = 0; i <bricks_in_base; i + +)drawbricksinline (i);/畫笫ri行方塊的方法，傳入?yún)?shù)：行數(shù) privatevoid

3、 drawbricksinline(intline) intbricknumber = bricks_in_base - line; for(inti = 0; i<bricknumber; i+)grect brick = new grect(brick_widthf brick_height);brick setlocation(line*brick_width/2 + i*brick_widths(win_widt h-bricksin_basebrick_width)/2,win_height-(line+1)ick_height-1);add(brick);/窗口的大小priv

4、atestaticfinalin工dth = 500 ; privatestaticf inalint win_height = 200 ;/磚塊的參數(shù)pr i vat estatic final int bktck_f7zdth = 30 ;/寬度privatestaticfinalintbricklheight = 12;/高度privatestaticf inalintbrtcks_i丄base = 14 ;/最底部“勺塊數(shù)運(yùn)行結(jié)果小應(yīng)用程序査看器：pyramkitest.class三啟動(dòng)小應(yīng)用程序。當(dāng)改變最底部的塊數(shù)時(shí)，也能正確運(yùn)行:小應(yīng)用程序査看器：pyramidtest.class

5、已啟動(dòng)小應(yīng)用稈序.problem 2：程序代碼/* file: targettest.java * assignment: assignment 2* description : ie 個(gè)靶子* author: yujiaping* date: 2016-4-03_*/import javaawt*;import acmgraphics*;import acmprogram*;publiculasstargettestextends graphicsprogram publicvoid run ()/設(shè)置窗口大小setsize ( win_widthf win_height);/畫3個(gè)實(shí)心圓形

6、成一個(gè)靶子圖案drawoval(100z midpoint r color.red);drawoval(65, midpoint r color.white); drawoval(35 a midpoint f color.red);/畫實(shí)心圓的函數(shù)，第一個(gè)參數(shù)表示直徑，第二個(gè)參數(shù)表示位置，第三個(gè)參數(shù)表示 是否為紅色privatevoid drawoval(doubled, point posz color color) goval oval = new goval(dr d);oval.setlocetion(posgetx()-d/2, posgety() -d/2 );oval.setf

7、illed(true);oval.setco丄or(color);oval.setfi丄丄co丄or(color);add(oval);/窗口的大小privatestaticfinalint win_width = 400; privatesta ti cf2.na.li.ntwin_height = 300;/屏幕的中點(diǎn)privatestaticfinal point midpoint = new point(win_width/2fwin_height/2);運(yùn)行結(jié)果 小應(yīng)用程序査看器：targettest.class三啟動(dòng)小應(yīng)用程序。增大靶了吋的效果:小應(yīng)用程序查番器：targettes

8、t.classp啟動(dòng)小應(yīng)用稈序-problem 3:程序代碼/* file: programhierarchy.java * assignment: assignment 2* description:畫一個(gè)關(guān)系圖* author: yujimping* date: 2016-4-03 import acmgraphics*; import gram.*;publiuulassprogramhierarchytestextends graphicsprogram publicvoid run ()/設(shè)置窗口大小setsize ( windows_widthf windows_he

9、ight);/畫4個(gè)帶字符串的框drawrectwithtext(rect_widthf rect_heightfwindows.width/2-rect_width/2f windows_height/2- vretical_dis/2-rect_heightf "program");drawrectwithtext(rect_widthf rect_height,windows_ wid th/ 2-rect_width/ 2 3-horizental_dis f windowsheight/2 + vretical_dis/2f ,graphicsprogramn )

10、;drawrectwithtext(rect_widthf rect_heightfwindows_ wid th/ 2-rect_width/ 2 , windowsheight/2 + vretical_dis/2f consoleprogram1');drawrectwithtext(rect_widthf rect_heightf windows_width/2+rect_width/2+horizental_disf windows_height/2 + vretical_dis/2r "dialogprogram");/畫3條線gline 1inel =

11、 newgline ( windows_width/2-horizental_dis-rect_widthr windowsheight/ 2 + vretical_dis/ 2 zwindows_ width/ 2 z windows_height/ 2 - vretical_dis/ 2 );add(linel);gline line2 = new gline(windowswidth/2 rwindows_height/ 2 + vretical_dis/ 2 r windows_ wid th/ 2 ,windows_height/ 2 - vretical_dis/ 2 );add(

12、line2);gline 1ine3 = newgline ( windows_width/2 + horizental_dis-rect_widthr windows_height/ 2 + vretical_dis/2 ,windows_ wid th/ 2 r windows_height/ 2 - vretical_dis/ 2 );add(line3);/畫帶字符串的框，字符串自動(dòng)居中。參數(shù)1, 2表示框的寬和高，參數(shù)3, 4表示 位置，參數(shù)5表示框屮的字符串privatevoid drawrectwithtext(intwidtha intheight, intpositioint

13、positiony, string str) grect rect= new grect(positionx, positiony, width, height); add (rect);glabel label = new glabel (str); doublew = labelgetwidth(); doubleh = labelgetascent();w = (rect_width - w) / 2 + positionx;h = (rect_height - h) / 2 + h + positiony; label.setlocmtion (w, h);add(label);/窗口

14、的寬和高privatestaticfinalintw工ndows_wtdth = 600; privatestaticfinalintwindows_height = 400;/方框的寬和高privatestaticfinalintrect_width = 120; privatestaticfinalintrecilheight = 4 0;/中間兩個(gè)框的距離privatestaticf inalint vreiticaldzs = 50 ; privatest at ic final inthokizentai»_dis = 50;l運(yùn)行結(jié)果 小應(yīng)用程厚査看器：programhi

15、erarchytest.class三啟動(dòng)小應(yīng)用程序。當(dāng)改變窗口大小或者是方框的寬高時(shí)，均正確運(yùn)行三啟動(dòng)小應(yīng)用程序.problem 4： 程序代碼/* file: sqrttest.java * assignment: assignment 2* description :計(jì)算3*3+bbfl勺平方flj* author: yujiaping* date: 2016-3-28_*/import acmprogram*;importacmutil*;publiculasssqrttestextends consoleprogram publicvoid run ()/提示信息print in (&

16、quot;計(jì)算 sqrt (a*a+b*b)”);/輸入3和binta = readlnt ( h aa :”)；intb = remdlnt (”請(qǐng)輸入b : ”)； /計(jì)算c的值 doublec = math.sqrt (a*a+b*b);/輸出結(jié)果ptin tin ( " sqrt ( n + a + " * '* + a + " z n 4- b+”*”+b+") = ”+c); 運(yùn)行結(jié)果寧波人學(xué)信息科學(xué)與工程學(xué)院小應(yīng)用程序查看器：sqrttestclass 計(jì)算 sqrt (a*a+b*b)請(qǐng)輸入a : 11請(qǐng)輸入b: 6sqrt(11

17、*11,6*6) = 12.529964086141668已啟動(dòng)小應(yīng)用程序(2)小應(yīng)用程序查舀器：sqrttest.class計(jì)算 sq rt (a*a+b*b)請(qǐng)輸入a : 3請(qǐng)輸入b : 4sqrt(3*3r4*4) = 5.0三啟動(dòng)小應(yīng)用程序。problem 5:程序代碼/* file: sqrttest.java* assignment: assignment 2* description :計(jì)算最大值和最小值* author: yujimping* date: 2016-3-28* /import acmprogram*;importacmut;publicclasssmallest

18、andlargsettestextends consoleprogram publicvoid run ()/提不信息 print in ("計(jì)算最大值和最小值*');intx = readlnt(h ;if(x = 0)printin (n你未輸入任何數(shù)據(jù)! ”)； return;intmax = x;intmin = x;while(x != signal) if(x>max) max = x; elseif(x<min) min = x;x = readlnt ( n ? n);/輸出最小值和最大值printin("sma丄丄est : h+min

19、); pr int in (11 largest : n+max);/標(biāo)記privatestaticf inalintszg!val= 0;運(yùn)行結(jié)果(1)小應(yīng)用程序查看器:? ? ? ? ? ? ?-計(jì)算遙大值和35小值一1117429-335esmallest : -3largest : 42已啟動(dòng)小應(yīng)用程序。(2)小應(yīng)用程序查看器:-計(jì)算讖大值和最小值 ? 0你未輸入任何數(shù)據(jù)！已啟動(dòng)小應(yīng)用程序。(3)小應(yīng)用程序查看器:-計(jì)算讖大值和最小值一? 2? 6smallest : 2largest : 2已啟動(dòng)小應(yīng)用程序。problem 6：程序代碼/* file: sqrttestjava 一一

20、一一一一一一一一一一一一一一一一一一一一一一* assignment: assignment 2系列的操作到1* description:讓一個(gè)數(shù)通過(guò)-* author: yu jiaping* date: 2016-3-28*/import gram*;importacme util.* ;publiuclassreachonetestextends consoleprogram publiuvoid run （）/提不信息print in （ n到達(dá) 1 ”）；/輸入整數(shù)nintn = readint（"請(qǐng)輸入一個(gè)整數(shù)n :"）;/定義步驟數(shù) intcou

21、nt = 0;/一直執(zhí)行直到n = 1 while（n != 1）/步驟數(shù)+ 1 count+； intnl = n;/如果n是偶數(shù)if（n % 2 = 0）n = nl / 2; printin （nl + n是偶數(shù)，所以除以2 : n+n）; else /如果n是奇數(shù)n=nl*3+l; printin （nl + ,f是奇數(shù)，所以嚴(yán)3 + 1 : h+n）;/輸出步驟數(shù)pri nt in （”一共用了”+count+”步至 lj達(dá)1 ”）；運(yùn)行結(jié)果（1）小應(yīng)用程序査看器:reacho. |n*3+l : 52 除以2 : 26 除以2 : 13n*3+l : 40除以2 : 20除以2 :

22、 10除以2 : 5到達(dá)i 請(qǐng)輸入一個(gè)整數(shù)n : 17 17是奇數(shù)，所以 52是偶數(shù)，所以 26是偶故，所以 13是奇數(shù)，所以 40是偶數(shù)，所以 20是偶數(shù)，所以 10是偶數(shù)，所以5是奇數(shù)，所以n*3*l : 16 16星偶數(shù)，所以除以2 : 8 8是偶數(shù)，所以除以2 : 4 4星偶數(shù)，所以除以2 : 2 2是偶數(shù)，所以除以2 : 1一共用了 12步到達(dá)1.已啟動(dòng)小應(yīng)用程序。 小應(yīng)用程序査看器：reach on etest.classsjii 1請(qǐng)輸入一個(gè)整數(shù)n : 1515是奇數(shù)，所以n*3-m : 4646是所以除以2 : 2323規(guī)奇故，所以"3“ : 7070果偶數(shù)，所以除以

23、2 : 3535是奇數(shù)，所以n*3+l : 106106泉偶故.所以嫁以2 : 5353是奇亂 所以n*3*l : 160160是偶數(shù)，所以除以2 : 8080是所以除以2 : 40所以除以2 : 2020是偶數(shù).所以除以2 : 1010sis數(shù)，所以除以2 : 55是奇數(shù).所以n*3>l : 16所以除以2 : 88泉偶數(shù).所以除以2 : 44杲偶數(shù)，所以除以2 : 22罡偶故.所以除以2 : 1 一共用了 17步到達(dá)1i己?jiǎn)?dòng)小應(yīng)用程序.錯(cuò)誤總結(jié)(1) 當(dāng)我把a(bǔ)cm的jar包復(fù)制到我的項(xiàng)冃中后，我發(fā)現(xiàn)寫程序時(shí)是 無(wú)法調(diào)用的，然后我查到需耍在項(xiàng)目的build path->configure bu