c语言课程设计设计报告

1、C语言课程设计报告 姓名： 学号： 年 月目录： 程序功能简介 题目分析 函数的调用关系图及主要算法 源程序及注释 测试流程 心得感想程序一：简单数学问题一、程序功能简介： 实现多个简单数学问题的求解。二、题目分析 完成与菜单项对应的功能设计1、 void FindNum( ); /*用穷举法找到被咬掉的数字*/2、 void FindRoot( ); /*求解方程ax2+bx+c=0的根*/3、void Detective( ); /*根据线索用穷举法找出牌照号码*/4、void Monkey( ); /*根据天数循环求出第一天所摘桃子数*/5、void Diamond( ); /*调用函数

2、Print_Diamond( )输出line行的钻石图案*/6、void Calculator( ); /*实现两个整数简单的加减乘除四则运算*/三、函数的调用关系图：menumenu_selectFindNumFindRootDetectiveMonkeyDiamondCalculatorPrint_Diamond四、源程序及注释：#include<stdio.h>#include<stdlib.h>#include<ctype.h>#include<math.h>int menu_select();void main() /*主函数*/ fo

3、r(;) switch(menu_select() case 1: FindNum(); system("pause"); break; case 2: FindRoot(); system("pause"); break; case 3: Detective(); system("pause"); break; case 4: Monkey(); system("pause"); break; case 5: Diamond(); system("pause"); break; case 6:

4、 Calculator(); system("pause"); break; case 0: printf("Goodbye!n"); system("pause"); exit(0); int menu_select() /*主函数菜单*/ char c; do system("cls"); printf("1.FindNumn"); printf("2.FindRootn"); printf("3.Detectiven"); printf("

5、4.Monkeyn"); printf("5.Diamondn"); printf("6.Calculatorn"); printf("0.Goodbye!n"); c=getchar(); while(c<'0'|c>'6'); return(c-'0');FindNum() /*找数字*/ int i; for(i=1;i<=9;i+) /*穷举*/ if(30+i)*6237=(10*i+3)*3564) printf("%dn",i

6、);FindRoot() /*找方程的根*/ float a10,b10,c10,disc,x1,x2,real,image; /*定义abc三个数组存放方程的系数*/ int i,k,j; for(i=0;i<10;i+) /*循环输入方程的三个系数*/ printf("please input three real numbers:n"); scanf("%f%f%f",&ai,&bi,&ci); printf("press '1' to continuenpress '2' t

7、o calculaten"); scanf("%d",&k); if (k=2) break; /*判断是否继续输入*/ for(j=0;j<=i;j+) printf("The equation %d",j+1); if (fabs(aj)<=1e-7) /*a=0的情况*/ printf(" is not quadratic, solution is %.2fn",-cj/bj); else disc=bj*bj-4*aj*cj; if (fabs(disc)<=1e-7) /*有两个相等的实根

8、的情况*/ printf(" is not quadratic,solution is: %.2fn",-bj/(2*aj); else if (disc>1e-7) /*有两个不等实根的情况*/ x1=(-bj+sqrt(disc)/(2*aj); x2=(-bj-sqrt(disc)/(2*aj); printf(" has distinct real roots: %.2f and %.2fn",x1,x2); else /*有两个虚根的情况*/ real=-bj/(2*aj); image=sqrt(-disc)/(2*aj); print

9、f(" has complex roots:n"); printf("%.2f+%.2f in",real,image); printf("%.2f-%.2f in",real,image); Detective() /*找牌照*/ int i,a,b,c,d;long m; for(i=32;i<=99;i+) /*穷举找出后四位号码*/ m=i*i; a=m%10; b=m/10%10; c=m/100%10; d=m/1000; if(a=b&&c=d) break; printf("%ldn&q

10、uot;,m+310000);Monkey() /*猴子吃桃问题*/ int n,i; long x; scanf("%d",&n); x=1; for(i=n-1;i>=1;i-) x=(x+1)*2; printf("%ldn",x);void print_diamond(int line) int i,j,k,m,n,p,q,a,b; for(i=1;i<=(49-line)/2;i+) /*输出(49-line)/2行的空格*/ printf("n"); for(j=1;j<=(line+1)/2;j

11、+) for(m=1;m<=(80-line)/2;m+) /*先输出(80-line)/2个空格*/ printf(" "); for(a=1;a<=(line-(2*j-1)/2;a+) /*行数j从1到中间一行输出(line-(2*j-1)/2个空格*/ printf(" "); for(n=1;n<=2*j-1;n+) /*中间行及上面每行输出2*j-1个"*"*/ printf("*"); printf("n"); for(k=1;k<=(line-1)/2;k

12、+) /*同理输出中间行下面的钻石图案*/ for(p=1;p<=(80-line)/2;p+) printf(" "); for(b=1;b<=k;b+) printf(" "); for(q=1;q<=line-2*k;q+) printf("*"); printf("n"); Diamond() /*钻石*/ int line; scanf("%d",&line); print_diamond(line); /*调用print_diamond输出钻石图案*/Cal

13、culator() /*简单计算器*/ int a4,b4,c4,i,j,k; /*定义ab两个数组存放数*/ char ch4; /*定义c存放运算符*/ FILE *fp1,*fp2; fp1=fopen("expres.txt","r"); /*打开expres文件*/ if(fp1=NULL)/*判断打开成功*/ printf("Can not open n"); exit(1); for(i=0;i<=3;i+) fscanf(fp1,"%d%c%d",&ai,&chi,&b

14、i);/*循环读入四道题*/ for(j=0;j<=3;j+) switch(chj)/*判断运算符*/ case '+':cj=aj+bj; break; case '-':cj=aj-bj; break; case '*':cj=aj*bj; break; case '/':cj=aj/bj; break; fp2=fopen("result.txt","w");/*打开文件result*/ if(fp2=NULL)/*判断打开文件是否成功*/ printf("Can n

15、ot open n"); exit(2); for(k=0;k<=3;k+)/*循环向文件中输入结果*/ fprintf(fp2,"%d%c%d=%dn",ak,chk,bk,ck); printf("succeed!n");五、测试流程：运行程序屏幕显示：1.FindNum 2.FindRoot 3.Detective 4.Monkey 5.Dismond 6.Calculator 0.Goodbye!输入：1回车显示：9 Press any key to continue 按任意键返回菜单输入：2回车显示：please input t

16、hree real number:输入：0回车 1回车 2回车显示：press 1 to continue press2 to calculate输入：1回车显示：please input three real number:(按照以上方法再输入三组数据：1，2，1；1，3，2；1，0，4后选择2，回车)显示：The equation 1 is not quadratic, solution is -2.00 The equation 2 is not quadratic, solution is -1.00 The equation 3 has distinct real roots: -1

17、.00and-2.00 The eauation 4 has complex roots: -0.00+2.00 i -0.00-2.00 i Press and key to continue 按任意键返回菜单输入：3 回车显示：317744 Press and key to continue 按任意键返回菜单输入：4 回车 10 回车显示：1534 Press and key to continue 按任意键返回菜单输入：5 回车 5 回车输出： * * * * * Press and key to continue 按任意键返回菜单输入：6 回车显示：succeed! Press and

18、 key to continue 按任意键返回菜单输入：0 回车显示：Goodbye! Press and key to continue 按任意键结束程序程序二：学生成绩简单管理一、 程序功能简介： 用结构体数组结构实现简单的学生管理成绩管理功能，要求具有数据输入、数据输出、数组排序、元素插入、元素删除、查询等功能。二、 题目分析完成与菜单项对应的功能设计1、int Input(struct student stu,int n); /*从键盘输入若干条纪录,依次存放到结构体数组stud中,n为数组原有纪录数,程序运行初始纪录数为0,函数返回纪录数.*/2、Display(struct stu

19、dent stu,int n); /*显示所有记录,每10个暂停一下,n为数组元素个数*/3、Sort_by_Name(struct student stu,int n); /*按姓名作升序排序,要求用选择排序法*/4、 int Insert_a_record(struct student stu,int n); /*输入待插入的姓名和成绩,按姓名作有序插入,输出插入成功信息,返回纪录个数*/5、int Delete_a_record(struct student stu,int n); /*输入待删除的姓名,经确认后删除该姓名的纪录,输出删除成功与否的信息.返回数组中的纪录数*/6、 int

20、 Query_a_recored(struct student stu,int n);/*输入待查找的姓名,查找该姓名的纪录,输出查找成功与否的信息和该学生的信息*/7、int Addfrom Text(struct student stud,int n); /*从文件添加一批纪录到数组中 ,返回添加纪录后的新的纪录数*/8、Write to Text(struct student stu,int n); /*将数组中的纪录全部写入文件records.txt*/三、函数调用关系图：mainmenu_selectInputDisplaySort_by_NameInsert_a_recordDel

21、ete_a_recordAddfromTextQuery_a_recordWritetoTextInsertDeleteSort_by_nameQueryInput_one_record四、源程序及注释：#include<stdio.h>#include<stdlib.h>#include<ctype.h>#include<string.h>typedef struct student /*定义结构体数组*/ char name20; float score;int menu_select();struct student Input_one_r

22、ecord() /*输入一个学生的成绩记录*/ struct student stud; printf("please input the name:n"); scanf("%s",&); printf("please input the score:n"); scanf("%f",&stud.score); return stud;int Input(struct student stu,int n) /*输入学生成绩记录*/ for(n=0;n<=39;n+) stun=Input_one

23、_record(); /*调用函数Input_ont_record输入学生成绩记录*/ printf("do you want to continue:('Y'or'N')"); getchar(); if(toupper(getchar()='N') break; return n+1;void Display(struct student stu, int n) /*输出所有学生的成绩记录*/ int i,j; for(i=0;i<3;i+) /*每屏显示十个记录然后暂停*/ system("cls&quo

24、t;); for(j=10*i;j<=10*i+9;j+) /*循环输出十个记录*/ if(j>n-1) break; printf("%s%20fn",,stuj.score); if(j>n-1) break; printf("press any key to display the next ten records:n"); system("pause"); void Sort_by_Name(struct student stu,int n) /*排序*/ int i,j,p; struct

25、 student t; for(i=0;i<=n-2;i+) /*选择排序法*/ p=i; for(j=i+1;j<=n-1;j+) /*找到最小数据的下标*/ if(strcmp(,)<0) p=j; if(p!=i) /*判断最小数据是否在第一个*/ t=stup;stup=stui;stui=t; printf("succeed!n");int Insert_a_record(struct student stu,int n) /*插入一个学生成绩记录*/ struct student new; struct st

26、udent *s; printf("please input a name:n"); scanf("%s",&); printf("please input the score:n"); scanf("%f",&new.score); s=&new; n=Insert(stu,n,s); /*调用函数Insert作有序插入*/ return n;int Insert(struct student stu,int n,struct student *s) /*有序插入*/ int i,j; f

27、or(i=0;i<=n-1;i+) if(strcmp(,s->name)>0) break; for(j=n-1;j>=i;j-) stuj+1=stuj; stui=*s; return n+1;int Delete_a_record(struct student stu,int n) /*删除一个学生的成绩记录*/ int k; struct student name_to_delete; struct student *name; printf("please input the name you want to delete:n&q

28、uot;); scanf("%s",&name_t); name=&name_to_delete; printf("Delete %s's record('Y'or'N'):n",name_t); getchar(); if(toupper(getchar()='Y') k=Delete(stu,n,name); /*调用函数Delete查找数据，若存在做删除*/ if(k=-1) /*判断是否存在*/ printf("The record does not exist!n

29、"); else n=k; return n;int Delete(struct student stu,int n,struct student *name) int i,j; for(i=0;i<=n;i+) if(strcmp(,name->name)=0) break; /*找到需要删除的数据下标i*/ if(i=n+1) /*判断要删除的数据是否存在*/ return -1; /*不存在，返回提示信息*/ else /*存在，做删除*/ for(j=i;j<=n;j+) stuj=stuj+1; printf("succeed!

30、n"); return n-1; void Query_a_record(struct student stu,int n) /*查找一个学生的成绩记录*/ struct student *name; struct student name_to_query; printf("please input the name you want to query:n"); scanf("%s",&name_t); name=&name_to_query; if(Query(stu,n,name)=-1) /*调用函数Query判断要查找

31、的记录是否存在*/ printf("%s does not exist!n"); /*不存在输出提示信息*/ else /*存在，输出记录*/ name_to_query=stuQuery(stu,n,name); printf("succeed!n%st%fn",name_t,name_to_query.score); int Query(struct student stu,int n,struct student *name) /*查找记录*/ int i; for(i=0;i<=n-1;i+) /*找到需要查找的记录的下标*/ if(str

32、cmp(,name->name)=0) break; if(i<=n-1) /*判断要查找的记录时候存在*/ return i; /*存在，返回下标i*/ else return -1; /*不存在，返回提示信息*/int AddfromText(struct student stu,int n) /*从文件中整批输入*/ int i,num; FILE *fp; /*定义文件指针*/ char 20; printf("Input the :n"); scanf("%s",); if(fp=fopen(,"r&qu

33、ot;)=NULL) printf("cann't open the file!n"); system("pause"); return n; fscanf(fp,"%d",&num); for(i=0;i<num;i+) /*循环读入文件内容*/ fscanf(fp,"%s%f",&stun+,&stun+i.score); n+=num; fclose(fp); Sort_by_Name(stu,n); /*调用函数Sort_by_Name排序*/ return

34、 n;void WritetoText(struct student stu,int n) /*将记录写到文件*/ int i; FILE *fp; char 20; printf("Write the records to a Text Filen"); printf("Input the :"); scanf("%s",); if(fp=fopen(,"w")=NULL) printf("cann't open the filen"); system("pause"

35、;); return; fprintf(fp,"%dn",n+1); for(i=0;i<=n-1;i+) /*循环将记录写到文件中*/ fprintf(fp,"%st%fn",,stui.score); fclose(fp); printf("succeed!n");void Quit()void main() /*主函数*/ struct student stu40; int n=0; for(;) switch(menu_select() case 1:n=Input(stu,n);system(&quo

36、t;pause");break; case 2:Display(stu, n);system("pause");break; case 3:Sort_by_Name(stu,n);system("pause");break; case 4:n=Insert_a_record(stu,n);system("pause");break; case 5:n=Delete_a_record(stu,n);system("pause");break; case 6:Query_a_record(stu,n);sys

37、tem("pause");break; case 7:n=AddfromText(stu,n);system("pause");break; case 8:WritetoText(stu,n);system("pause");break; case 0:Quit();system("pause"); exit(0); int menu_select() /*显示主菜单*/ char c; do system("cls"); printf("1.Input Recordsn")

38、; printf("2.Display()n"); printf("3.Sort by Namen"); printf("4.Insert a recordn"); printf("5.Delete a recordn"); printf("6.Queryn"); printf("7.AddfromTextn"); printf("8.WritetoTextn"); printf("0.Goodbye!n"); printf("

39、;Give your choice (1-8,0):"); c=getchar(); while(c<'0'|c>'8'); return(c-'0');五、测试流程：运行程序屏幕显示：1.Input Records 2.Display() 3.Sort by Name 4.Insert a record 5.Delete a record 6.Query 7.AddfromText 8.WritetoText 0 .Goodbye! Give your choice (1-8,0):输入：1 回车显示：please inp

40、ut the name: 输入：hhm 回车显示：please input the score：输入：75 回车显示：Do you want to continue: (YorN)输入：y 回车（按照上述方法输入三组学生记录：gm，95；zy，95；zbn，95 再选择N）屏幕显示：press any key to continue 按任意键返回主菜单输入：2 回车显示：hhm 75.000000 gm 95.000000 zy 95.000000 zbn 95.000000 Press any key to continue按任意键返回主菜单输入：3回车显示：succeed！ Press a

41、ny key to continue按任意键返回主菜单输入：2回车显示：gm 95.000000 hhm 75.000000 zbn 95.000000 zy 95.000000 Press any key to continue按任意键返回主菜单输入：4 回车显示：please input a name: 输入：xjj回车显示：please input the score: 输入：88回车显示：Press any key to continue按任意键返回主菜单输入：2 回车显示：gm 95.000000 hhm 75.000000 xjj 88.000000 zbn 95.000000 zy