云南大学 软件学院 大学物理实验5

云南大学软件学院 实验报告课程： 大学物理实验 学期： 任课教师： 班级： = 学号： == 序号： = 姓名： = 成绩：实验6 简谐振动的合成内容一 同方向同频率简谐振动的合成两个简谐振动的方程为使用matlab 编写程序，求x1,x2,合振动的波形，讨论相位差对合成振动的影响。

相位差至少讨论4中情况（1.为0°;2.为180°；3.小于180°；4.大于180°），要求所有波形画在同一个figure 中。

()()⎩⎨⎧+=+=222111cos cos ϕωϕωt A x t A x内容二 相互垂直方向同频率简谐振动的合成两个简谐振动的方程为使用matlab 编写程序，求x,y,合振动的波形，讨论相位差对合成振动的影响。

相位差至少讨论4中情况（1.为0°;2.为180°；3.小于180°；4.大于180°），要求所有波形画在同一个figure 中。

()()⎩⎨⎧+=+=y y x x t A y t A x ϕωϕωcos cos内容三相互垂直方向不同频率简谐振动的合成(李萨如图形) 使用matlab编写程序，画李萨如图形，要求：1.至少4种频率比2.至少8种相位差3.所有图形画在同一个figure中，添加标注。

如：cleart = 0:0.01:4;Ax = 1;Ay = 3;w1 = 1; w2 = 1./2;w3 = 2./3;w4 = 3./4;w5 = 2./5;m0 = 0;m1 = 0;m2 = pi./4;m3 = pi./2;m4 = 3.*pi./4;m5 = pi;m6 = 5.*pi./4; m7 =3.*pi./2;m8 = 7.*pi./4; x0 = Ax.*cos(2.*pi*t+m0);y11 = Ay.*cos(2.*w1.*pi*t+m1);y12 = Ay.*cos(2.*w1.*pi*t+m2);y13 = Ay.*cos(2.*w1.*pi*t+m3);y14 = Ay.*cos(2.*w1.*pi*t+m4);y15 = Ay.*cos(2.*w1.*pi*t+m5);y16 = Ay.*cos(2.*w1.*pi*t+m6);y17 = Ay.*cos(2.*w1.*pi*t+m7);y18 = Ay.*cos(2.*w1.*pi*t+m8);y21 = Ay.*cos(2.*w2.*pi*t+m1);y22 = Ay.*cos(2.*w2.*pi*t+m2);y23 = Ay.*cos(2.*w2.*pi*t+m3);y24 = Ay.*cos(2.*w2.*pi*t+m4);y25 = Ay.*cos(2.*w2.*pi*t+m5);y26 = Ay.*cos(2.*w2.*pi*t+m6);y27 = Ay.*cos(2.*w2.*pi*t+m7);y28 = Ay.*cos(2.*w2.*pi*t+m8);y31 = Ay.*cos(2.*w3.*pi*t+m1);y32 = Ay.*cos(2.*w3.*pi*t+m2);y33 = Ay.*cos(2.*w3.*pi*t+m3);y34 = Ay.*cos(2.*w3.*pi*t+m4);y35 = Ay.*cos(2.*w3.*pi*t+m5);y36 = Ay.*cos(2.*w3.*pi*t+m6);y37 = Ay.*cos(2.*w3.*pi*t+m7);y38 = Ay.*cos(2.*w3.*pi*t+m8);y41 = Ay.*cos(2.*w4.*pi*t+m1);y42 = Ay.*cos(2.*w4.*pi*t+m2);y43 = Ay.*cos(2.*w4.*pi*t+m3);y44 = Ay.*cos(2.*w4.*pi*t+m4);y45 = Ay.*cos(2.*w4.*pi*t+m5);y46 = Ay.*cos(2.*w4.*pi*t+m6);y47 = Ay.*cos(2.*w4.*pi*t+m7);y48 = Ay.*cos(2.*w4.*pi*t+m8);y51 = Ay.*cos(2.*w5.*pi*t+m1);y52 = Ay.*cos(2.*w5.*pi*t+m2);y53 = Ay.*cos(2.*w5.*pi*t+m3);y54 = Ay.*cos(2.*w5.*pi*t+m4);y55 = Ay.*cos(2.*w5.*pi*t+m5);y56 = Ay.*cos(2.*w5.*pi*t+m6);y57 = Ay.*cos(2.*w5.*pi*t+m7);y58 = Ay.*cos(2.*w5.*pi*t+m8);subplot(5,8,1);plot(x0,y11);Axis([-4 4 -4 4]);text(-16,0,'ω =1');text(-2,7,'0');text(-16,7,'相位差');subplot(5,8,2);plot(x0,y12);Axis([-4 4 -4 4]);text(-2,7,'π/4'); subplot(5,8,3);plot(x0,y13);Axis([-4 4 -4 4]);text(-2,7,'π/2'); subplot(5,8,4);plot(x0,y14);Axis([-4 4 -4 4]);text(-2,7,'3π/4'); subplot(5,8,5);plot(x0,y15);Axis([-4 4 -4 4]);text(-2,7,'π');subplot(5,8,6);plot(x0,y16);Axis([-4 4 -4 4]);text(-2,7,'5π/4'); subplot(5,8,7);plot(x0,y17);Axis([-4 4 -4 4]);text(-2,7,'3π/2'); subplot(5,8,8);plot(x0,y18);Axis([-4 4 -4 4]);text(-2,7,'7π/4'); subplot(5,8,9);plot(x0,y21);Axis([-4 4 -4 4]);text(-16,0,'ω = 1/2'); subplot(5,8,10);plot(x0,y22);Axis([-4 4 -4 4]);subplot(5,8,11);plot(x0,y23);Axis([-4 4 -4 4]);subplot(5,8,12);plot(x0,y24);Axis([-4 4 -4 4]);subplot(5,8,13);plot(x0,y25);Axis([-4 4 -4 4]);subplot(5,8,14);plot(x0,y26);Axis([-4 4 -4 4]);subplot(5,8,15);plot(x0,y27);Axis([-4 4 -4 4]);subplot(5,8,16);plot(x0,y28);Axis([-4 4 -4 4]);subplot(5,8,17);plot(x0,y31);Axis([-4 4 -4 4]);text(-16,0,'ω = 2/3'); subplot(5,8,18);plot(x0,y32);Axis([-4 4 -4 4]);subplot(5,8,19);plot(x0,y33);Axis([-4 4 -4 4]);subplot(5,8,20);plot(x0,y34);Axis([-4 4 -4 4]);subplot(5,8,21);plot(x0,y35);Axis([-4 4 -4 4]);subplot(5,8,22);plot(x0,y36);Axis([-4 4 -4 4]);subplot(5,8,23);plot(x0,y37);Axis([-4 4 -4 4]);subplot(5,8,24);plot(x0,y38);Axis([-4 4 -4 4]);subplot(5,8,25);plot(x0,y41);Axis([-4 4 -4 4]);text(-16,0,'ω = 3/4'); subplot(5,8,26);plot(x0,y42);Axis([-4 4 -4 4]);subplot(5,8,27);plot(x0,y43);Axis([-4 4 -4 4]);subplot(5,8,28);plot(x0,y44);Axis([-4 4 -4 4]);subplot(5,8,29);plot(x0,y45);Axis([-4 4 -4 4]);subplot(5,8,30);plot(x0,y46);Axis([-4 4 -4 4]);subplot(5,8,31);plot(x0,y47);Axis([-4 4 -4 4]);subplot(5,8,32);plot(x0,y48);Axis([-4 4 -4 4]);subplot(5,8,33);plot(x0,y51);Axis([-4 4 -4 4]);text(-16,0,'ω = 2/5'); subplot(5,8,34);plot(x0,y52);Axis([-4 4 -4 4]);subplot(5,8,35);plot(x0,y53);Axis([-4 4 -4 4]);subplot(5,8,36);plot(x0,y54);Axis([-4 4 -4 4]);subplot(5,8,37);plot(x0,y55);Axis([-4 4 -4 4]);subplot(5,8,38);plot(x0,y56);Axis([-4 4 -4 4]);subplot(5,8,39);plot(x0,y57);Axis([-4 4 -4 4]);subplot(5,8,40);plot(x0,y58);Axis([-4 4 -4 4]);。

实验五 80x86分支，循环，子程序，宏程序设计姓名：学号：序号：班级：分数：1．编写设计一个2，10，16进制的码制转换程序，要求：①程序有友好的运行界面；②输入任意类型进制的数据，转换为其余两种进制类型；③要求程序有方便的输入输出功能；④要求程序能够处理基本的错误信息；⑤程序的基本结构采用子程序结构；源程序：.8086.model small.datastr db 0ah,0dh,'please input 1 to 8 to choise the function',0ah,0dhdb '1.btd',0ah,0dhdb '2.bth',0ah,0dhdb '3.dth',0ah,0dhdb '4.dtb',0ah,0dhdb '5.htb',0ah,0dhdb '6.htd',0ah,0dhdb '7.again',0ah,0dhdb '8.end',0ah,0dh,'$'b db 'Please input a Binary number',0ah,0dh,'$'d db 'Please input a decimal number',0ah,0dh,'$'h db 'Please input a Hexadecimal number',0ah,0dh,'$'agai db 'Please input again',0ah,0dh,'$'wr db 'A wrong number!',0ah,0dh,'$'num db 6,?,6 dup (?)num1 db 16,?,16 dup (?).codestart:mov ax,@datamov ds,axchoice:lea dx,strmov ah,9int 21hmov ah,1int 21hjmp C1mov ah,2int 21hmov dl,0dhmov ah,2int 21hjmp choiceMo: mov dx,offset agai mov ah,9int 21hjmp choiceC1: cmp al,'1'jnz C2call btdjmp againC2: cmp al,'2'jnz C3call bthjmp againC3: cmp al,'3'jnz C4call dthjmp againC4: cmp al,'4'jnz C5call dtbjmp againC5: cmp al,'5'jnz C6call htbjmp againC6: cmp al,'6'jnz C7call htdjmp againC7: cmp al,'7'jnz C8jmp Mojnz wrongen: mov ah,4chint 21hwrong:lea dx,wrmov ah,9int 21hjmp Mobtd procmov dx,offset bmov ah,9int 21hmov dx,offset num1mov ah,0ahint 21hmov dl,0ahmov ah,2int 21hmov dl,0dhmov ah,2int 21hmov di,offset num1mov ch,[di+1]zq2: mov dl,[di+2]cmp dl,39hjg zqsub dl,30hjmp zq1zq: sub dl,37hzq1: mov dh,0or bx,dxcmp ch,1jz zq3rol bx,1zq3: inc didec chjnz zq2jj1: cmp bx,10000 jc jjsub bx,10000inc sijmp jj1jj: mov dx,siand dx,0fhadd dl,30hmov ah,2int 21hxor si,siL1: cmp bx,1000jc L2sub bx,1000inc sijmp L1L2: mov dx,siand dx,0fhadd dl,30hmov ah,2int 21hxor si,siL3: cmp bx,100jc L4sub bx,100inc sijmp L3L4: mov dx,siand dx,0fhadd dl,30hmov ah,2int 21hxor si,siL5: cmp bx,10jc L6sub bx,10inc sijmp L5L6: mov dx,simov ah,2int 21hxor dx,dxmov dx,bxand dx,0fhadd dl,30hmov ah,2int 21hretbtd endpbth procmov dx,offset bmov ah,9int 21hmov dx,offset num1mov ah,0ahint 21hmov dl,0ahmov ah,2int 21hmov dl,0dhmov ah,2int 21hmov di,offset num1mov ch,[di+1]K1: mov al,[di+2]sub al,30hor bl,alrol bx,1inc didec chjnz K1ror bx,1mov ch,4mov cl,4K4: rol bx,clmov dl,bland dl,0fhadd dl,30hjmp K3K2: add dl,37hK3: mov ah,2int 21hdec chjnz K4retbth endpdth proclea dx,dmov ah,9int 21hmov dx,offset nummov ah,0ahint 21hmov dl,0ahmov ah,2int 21hmov dl,0dhmov ah,2int 21hmov di,offset nummov cl,[di+1]add di,2mov dl,[di]sub dl,30hmov al,10mul dlmov ch,0dec cxI: inc dimov dl,[di]sub dl,30hmov dh,0add ax,dxcmp cx,1jz I1mov bx,axI1: loop Imov bx,axmov si,offset nummov ch,[si+1]inc chinc chmov cl,4O: rol bx,clmov dl,bland dl,0fhcmp dl,9jg MMadd dl,30hjmp PPMM: add dl,37h PP: mov ah,2int 21hdec chjnz Oretdth endpdtb proclea dx,dmov ah,9int 21hmov dx,offset nummov ah,0ahint 21hmov dl,0ahmov ah,2int 21hmov dl,0dhmov ah,2int 21hmov di,offset nummov cl,[di+1]add di,2mov al,10mul dlmov ch,0dec cxc: inc dimov dl,[di]sub dl,30hmov dh,0add ax,dxcmp cx,1jz ccmov bx,10mul bxmov bx,axcc: dec cxjnz cmov bx,axmov cx,16x: rol bx,1mov dl,bland dl,01hadd dl,30hmov ah,2int 21hloop xretdtb endphtb procmov dx,offset hmov ah,9int 21hmov dx,offset nummov ah,0ahint 21hmov dl,0ahmov ah,2int 21hmov dl,0dhmov ah,2int 21hmov ch,[di+1]Z1: mov bl,[di+2]cmp bl,39hjg ZZsub bl,30hjmp ZXZZ: sub bl,37hZX: mov cl,4mov bh,4shl bl,clZ: rol bl,1mov dl,bland dl,01hadd dl,30hY: mov ah,2int 21hdec bhjnz Zinc didec chjnz Z1rethtb endphtd procmov dx,offset hmov ah,9int 21hmov dx,offset nummov ah,0ahint 21hmov dl,0ahmov ah,2int 21hmov dl,0dhmov ah,2int 21hmov di,offset nummov ch,[di+1]u2: mov dl,[di+2]sub dl,30hjmp u1u: sub dl,37h u1: mov dh,0or bx,dxcmp ch,1jz Rmov cl,4rol bx,cl R: inc didec chjnz u2xor si,sis1: cmp bx,1000jc s2sub bx,1000inc sijmp s1s2: mov dx,siand dx,0fhadd dl,30hmov ah,2int 21hxor si,sis3: cmp bx,100jc s4sub bx,100inc sijmp L3xor si,sis4: cmp bx,10jc s5sub bx,10inc sijmp s3s5: mov dx,siand dx,0fhadd dl,30hmov ah,2int 21hmov dx,bxand dx,0fhadd dl,30hmov ah,2int 21hrethtd endpend start实验结果：。

课程：数据结构实验学期：2014-2015学年第一学期任课教师：专业：信息安全学号：姓名：成绩：实验5 图基础实验一、实验目的1.掌握图的存储结构及其遍历。

二、实验软硬件环境（CPU、OS、IDE）：三、实验任务（要求写出核心代码，并对运行结果截图）1）使用邻接矩阵和邻接表储表示分别实现如下给定的图1、图2、图3所示图的物理存储结构。

2）在1）所建立的图形存储结构上分别实现深度优先搜索遍历和广度优先搜索遍历，并给出遍历结果（序列）。

图3 有向图实验代码：#include<stdio.h>#include<stdlib.h>#define MAXVEX 20#define OK 1#define ERROR 0#define OVERFLOW -1#define INFINITY 65535#define QueueSize 20 //队列中最大元素个数typedef int QElemType; //队列的元素的类型typedef int VertexType;typedef int EdgeType;typedef enum{False,True}Boolean; //Boolean是布尔类型，其值是ture或false Boolean visited[MAXVEX]; //访问标志的数组。

typedef struct{VertexType vexs[MAXVEX];EdgeType arc[MAXVEX][MAXVEX];int numVertexes,numEdges;} MGraph; //邻接矩阵。

typedef struct EdgeNode //边表结点。

{int adjvex;struct EdgeNode *next;}EdgeNode;typedef struct VertexNode //顶点表结点。

{int data;EdgeNode *firstedge;}VertexNode,AdjList[MAXVEX];typedef struct{AdjList adjlist;int numVertexes,numEdges; //图中当前顶点数边数。

云南大学软件学院实验报告课程：数据库概论任课教师：包崇明姓名：学号：专业：成绩：实验5 数据查询-基于CAP数据库完成教科书第三章如下习题(SQL语句及结果截屏)：[3.2]b： select aid from AGENTS where [percent] >=all(select MAX([percent]) from AGENTS)截图：[3.5]：select aid,cid from AGENTS A,CUSTOMERS C where aid not in (select aid from ORDERS X where X.cid=C.cid and X.aid=a.aid) order by aid,cid截图：[3.8]a：注意：中文版教材翻译有误，应该是求每个顾客在他购买的所有产品中的最大花费值，参考英文原版select cid,pid,MAX（dollars）as MAXSPENTfrom ORDERSgroup by cid,pid,dollars截图：[3.8]b：select AVG (MAXSPENT) as AVERAGEfrom (select cid,pid,MAX(dollars) as MAXSPENTfrom ORDERSgroup by cid,pid,dollars)as x group by cid截图：[3.11]b：需要先求出按cid,pid分组的qty均值，并完整数据截图；其次，本题中最小qty均值修改为900select cid,pid,AVG(qty) as AVGQTYfrom ORDERSgroup by cid,pid,qty截图：select cid from(select cid,pid,AVG(qty) as AVGQTYfrom ORDERS group by cid,pid,qty)as x where AVGQTY > 900截图：[3.11]d： select aid from ORDERSwhere cid IN(select cidfrom CUSTOMERSwhere city = 'Duluth' or city = 'Kyoto')截图：[3.11]f： select pid from ORDERSwhere cid IN(select cidfrom CUSTOMERSwhere city = 'Dallas')截图：[3.11]h： delete from AGENTS where aname = 'Gray' 截图：INSERT INTO AGENTS(aid,aname,city,[percent]) values('a04','Gray','New york',6)截图：[3.11]j： update PRODUCTSset price = 1.1*pricewhere city = 'Duluth' or city = 'Dallas'截图：[3.11]k：select cid,sum(dollars) as totalDollars from ORDERS where aid = 'a04'group by cid截图：[3.11]l： select aid,[percent] from AGENTS a where not exists(select cid from CUSTOMERS cwhere city = 'Duluth' and not exists(select o.cid from ORDERS owhere o.cid = c.cid and a.aid = o.aid))order by [percent]截图：。