《數(shù)據(jù)結(jié)構(gòu)》實(shí)驗(yàn)報(bào)告抽象數(shù)據(jù)類型復(fù)數(shù)的實(shí)現(xiàn)

1、實(shí) 驗(yàn) 報(bào) 告學(xué) 院: 信息工程學(xué)院 專 業(yè): 計(jì)算機(jī) 信息工程學(xué)院計(jì)算機(jī)實(shí)驗(yàn)中心制數(shù)據(jù)結(jié)構(gòu)實(shí)驗(yàn)報(bào)告姓名學(xué)號(hào)日期2011.3.25 7-8節(jié)實(shí)驗(yàn)室計(jì)算機(jī)基礎(chǔ)實(shí)驗(yàn)室指導(dǎo)教師張有華設(shè)備編號(hào)實(shí)驗(yàn)題目實(shí)驗(yàn)1 抽象數(shù)據(jù)類型復(fù)數(shù)的實(shí)現(xiàn)一 實(shí)驗(yàn)內(nèi)容實(shí)驗(yàn)1 抽象數(shù)據(jù)類型復(fù)數(shù)的實(shí)現(xiàn)二 實(shí)驗(yàn)?zāi)康?. 了解抽象數(shù)據(jù)類型（adt）的基本概念，及描述方法。2. 通過(guò)對(duì)復(fù)數(shù)抽象數(shù)據(jù)類型adt的實(shí)現(xiàn)，熟悉c語(yǔ)言語(yǔ)法及程序設(shè)計(jì)。為以后章節(jié)的學(xué)習(xí)打下基礎(chǔ)。三 需求分析復(fù)數(shù)抽象數(shù)據(jù)類型adt的描述及實(shí)現(xiàn)。 復(fù)數(shù)adt的描述 adt complex 數(shù)據(jù)對(duì)象：d= c1,c2 c1,c2floatset 數(shù)據(jù)關(guān)系：r= c1, c

2、2 d 基本操作：創(chuàng)建一個(gè)復(fù)數(shù) initcomplex(); 輸出一個(gè)復(fù)數(shù) outcomplex(); 求兩個(gè)復(fù)數(shù)相加之和 addcomplex(); 求兩個(gè)復(fù)數(shù)相減之差 subcomplex(); 求兩個(gè)復(fù)數(shù)相乘之積 mulcomplex(); 求兩個(gè)復(fù)數(shù)的商 scomplex(); 等等; adt complex;本實(shí)驗(yàn)實(shí)現(xiàn)使用tc2.0實(shí)現(xiàn)復(fù)數(shù)的描述及操作。具體實(shí)現(xiàn)要求：1從鍵盤分別輸入2個(gè)復(fù)數(shù)，并可修改已輸入的復(fù)數(shù)。2能輸出指定的復(fù)數(shù)。3兩個(gè)復(fù)數(shù)相加之和，觀察輸出結(jié)果。4兩個(gè)復(fù)數(shù)相加之差，觀察輸出結(jié)果。5求兩個(gè)復(fù)數(shù)相乘之積，觀察輸出結(jié)果。6.求兩個(gè)復(fù)數(shù)的商，觀察輸出結(jié)果。7用戶可看到如

3、下界面： * * 1.輸入復(fù)數(shù)c1 * * 2.輸入復(fù)數(shù)c2 * * 3.輸出復(fù)數(shù)c1 * * 4.輸出復(fù)數(shù)c2 * * 5.求c1和c2的和 * * 6.求c1和c2的差 * * 7.求c1和c2的積 * * 8.求c1和c2的商 * * 0.結(jié)束 * * 四 詳細(xì)設(shè)計(jì)步驟1：復(fù)數(shù)的抽象數(shù)據(jù)類型的定義。adt complex 數(shù)據(jù)對(duì)象：d=a,b|a,bfloatset 數(shù)據(jù)關(guān)系：r= |a,b d 基本操作：initcomplex(&c,vr,vi);操作結(jié)果：構(gòu)造一個(gè)復(fù)數(shù),元素a,b分別被賦以參數(shù)vr,vi的值。outcomplex(c);操作結(jié)果:輸出一個(gè)復(fù)數(shù)。addcomplex(&

4、c,c1,c2);操作結(jié)果：求兩個(gè)復(fù)數(shù)c1,c2之和,結(jié)果存入c。subcomplex(&c,c1,c2); 操作結(jié)果：求兩個(gè)復(fù)數(shù)c1，c2之差，結(jié)果存入c。mulcomplex(&c,c1,c2); 操作結(jié)果：求兩個(gè)復(fù)數(shù)c1，c2的積，結(jié)果存入c。 adt complex步驟2：復(fù)數(shù)的存儲(chǔ)結(jié)構(gòu)及相關(guān)操作的聲明。復(fù)數(shù)的存儲(chǔ)結(jié)構(gòu)#include #include typedef struct complexfloat real;float image;complex;基本操作的聲明：status initcomplex(&c,vr,vi);操作結(jié)果：構(gòu)造一個(gè)復(fù)數(shù),元素a,b分別被賦以參數(shù)vr,v

5、i的值。void outcomplex(c);操作結(jié)果:輸出一個(gè)復(fù)數(shù)。complex addcomplex(&c,c1,c2);操作結(jié)果：求兩個(gè)復(fù)數(shù)c1,c2之和,結(jié)果存入c。complex subcomplex(&c,c1,c2); 操作結(jié)果：求兩個(gè)復(fù)數(shù)c1，c2之差，結(jié)果存入c。complex mulcomplex(&c,c1,c2); 操作結(jié)果：求兩個(gè)復(fù)數(shù)c1，c2的積，結(jié)果存入c。complex scomplex(&c,c1,c2);操作結(jié)果：求兩個(gè)復(fù)數(shù)c1，c2的商，結(jié)果存入c。步驟3：復(fù)數(shù)的基本操作的實(shí)現(xiàn)。復(fù)數(shù)的基本操作：/構(gòu)造一個(gè)復(fù)數(shù),元素a,b分別被賦以參數(shù)vr,vi的值sta

6、tus initcomplex(complex &c, float vr, float vi) c.real = vr; c.image =vi; return ok;/輸出一個(gè)復(fù)數(shù)void outcomplex(complex c) printf(c.real“+”c.image“i”);/求兩個(gè)復(fù)數(shù)c1,c2之和,結(jié)果存入ccomplex addcomplex(complex &c,complex c1,complex c2); c.real = c1.real + c2.real; c.image = c1.image + c2.image; return c; /求兩個(gè)復(fù)數(shù)c1，c2之

7、差，結(jié)果存入ccomplex subcomplex(complex &c,complex c1,complex c2) c.real = c1.real - c2.real; c.image = c1.image - c2.image; return c; /求兩個(gè)復(fù)數(shù)c1，c2的積，結(jié)果存入ccomplex mulcomplex(complex &c,complex c1,complex c2) c.real = c1.real * c2.real + c1.image * c2.image; c.image = c1.real * c2.image + c1.image * c2.real

8、; return c; /求兩個(gè)復(fù)數(shù)c1，c2的商，結(jié)果存入ccomplex scomplex(complex &c, complex c1, complex c2)c.real = (c1.real * c2.real + c1.image * c2.image)/(c2.real * c2.real + c2.image * c2.image); c.image = (c1.image * c2.real - c1.real * c2.image)/(c2.real * c2.real + c2.image * c2.image);return c;步驟4：上機(jī)編程與調(diào)試 #include

9、 stdafx.h#include complex0515.h#include user.hint main(int argc, char* argv)int flag,flag1;float cr,ci,vr,vi;complex c1,c2,c; ccomplex0506 c;printf( * n);printf( * 1.輸入復(fù)數(shù)c1 * n);printf( * 2.輸入復(fù)數(shù)c2 * n);printf( * 3.輸出復(fù)數(shù)c1 * n);printf( * 4.輸出復(fù)數(shù)c2 * n);printf( * 5.求c1和c2的和 * n);printf( * 6.求c1和c2的差 * n

10、);printf( * 7.求c1和c2的積 * n);printf( * 8.求c1和c2的商 * n); printf( * 0.結(jié)束 * n); printf( * n);while(1)printf(請(qǐng)輸入您的選擇（08）:);scanf(%d,&flag);switch(flag) case 1: printf(請(qǐng)分別輸入復(fù)數(shù)c1的實(shí)部和虛部（空格隔開）:); scanf(%f %f,&vr,&vi); break; case 2: printf(請(qǐng)分別輸入復(fù)數(shù)c2的實(shí)部和虛部（空格隔開）:); scanf(%f %f,&cr,&ci); break; case 3: c.initc

11、omplex(c1,vr,vi); printf(c1=); c.outcomplex(c1); /復(fù)數(shù)的初始化 break; case 4: c.initcomplex(c2,cr,ci); printf(c2=); c.outcomplex(c2); break; case 5: c.addcomplex(c,c1,c2); /求兩個(gè)復(fù)數(shù)的和 printf(c1+c2=); c.outcomplex(c); break; case 6: c.subcomplex(c,c1,c2); /求兩個(gè)復(fù)數(shù)的差 printf(c1-c2=); c.outcomplex(c); break; case

12、7: c.mulcomplex(c,c1,c2); /求兩個(gè)復(fù)數(shù)的積 printf(c1*c2=); c.outcomplex(c); break; case 8: c.scomplex(c,c1,c2); /求兩個(gè)復(fù)數(shù)的商 printf(c1/c2=); c.outcomplex(c); break; case 0: printf(結(jié)束n); flag1=1; break; default: printf(輸入不合法！n); break;/switchif(flag1=1) break;/whilereturn 0;運(yùn)行結(jié)果：圖1.1開始定義變量 ccomplex0515 c;complex

13、 c1,c2,cadd,csub,cmul;實(shí)現(xiàn)各種復(fù)數(shù)運(yùn)算的算法打印出來(lái)結(jié)束步驟五:實(shí)驗(yàn)總結(jié)1. 通過(guò)本次實(shí)驗(yàn)，基本掌握抽象數(shù)據(jù)類型的定義方法及要求；2. 基本掌握c語(yǔ)言程序設(shè)計(jì)的規(guī)范操作流程；3. 編程過(guò)程中有些地方考慮不全面，程序不夠健壯，；ut2apodfxxc02gybkskcww97mrqqwhoj5tl15zt6jipyytycummtarp3v1n5luizi3xh3bhwyreko8d9g7nmzqowpjetldrw08gvs8dsdqqygc3ce7moo2tlf0jf1gk74iuxybmtivr97ckrfvqult5fn2t6mpjr6rbzvpsortzvij5n

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