c语言程序设计实例求C语言程序设计实例，要100行以上

求c语言程序设计实例80到100行

#include <stdio.h> /*头函数*/ int main(void) /*主函数*/ { char c; /*指定c成为字符变量*/ printf("请输入0到9的任意一个数字 "); /*输出请输入0到9的任意一个数字*/ c=getchar(); /*输入一个字符*/ while(c>=48&&c<=57) /*c的取值范围*/ { getchar(); switch(c) /*根据c的值转换*/ { case '0': printf("你喜欢奋斗吗? "); /*假如c=0,输出你喜欢奋斗吗?*/ break; /*中断跳出，执行c=getchar()*/ case '1': printf("你喜欢当模特吗? "); /*假如c=1,输出你喜欢当模特吗?*/ break; /*中断跳出，执行c=getchar()*/ case '2': printf("你喜欢和谐吗? "); /*假如c=2,输出你喜欢和谐吗?*/ break; /*中断跳出，执行c=getchar()*/ case '3': printf("你喜欢道家吗? "); /*假如c=3,输出你喜欢道家吗?*/ break; /*中断跳出，执行c=getchar()*/ case '4': printf("你是广东人吗? "); /*假如c=4,输出你是广东人吗?*/ break; /*中断跳出，执行c=getchar()*/ case '5': printf("你喜欢武术吗? "); /*假如c=5,输出你喜欢武术吗?*/ break; /*中断跳出，执行c=getchar()*/ case '6': printf("祝你一帆风顺! "); /*假如c=6,输出祝你一帆风顺!*/ break; /*中断跳出，执行c=getchar()*/ case '7': printf("观察等待好机会! "); /*假如c=7,输出观察等待好机会!*/ break; /*中断跳出，执行c=getchar()*/ case '8': printf("你喜欢交际吗? "); /*假如c=8,输出你喜欢交际吗?*/ break; /*中断跳出，执行c=getchar()*/ case '9': printf("追求完美吧! "); /*假如c=9,输出追求完美吧!*/ break; /*中断跳出，执行c=getchar()*/ } c=getchar(); /*再输入一个字符*/ } return 0; /*返回值*/ }

c语言程序设计案例。要100行左右的。。核心代码50行左右，可以在TC上运行的。拜谢了，，，

下面是个C++代码。

估计你会用上，呵呵。

~！~ #include "stdio.h" #include "stdlib.h" class Student { public: int id; int eng; int math; puter; double avg; double total; Student(){id=-1;eng=-1;;math=puter=-1;avg=-1;total=-1;}; double Aveage() { return avg=(double)(eng+puter)/(double)3; }; double Sum() { return total=(double)(eng+puter); }; }; void EX(Student*A,Student*B)//排序 { Student temp; temp=*A; *A=*B; *B=temp; } void PX(Student* A,int s) { Student* R=new Student[s]; for(int i=0;i<s;i++) { for(int j=i+1;j<s;j++) { if((A+i)->avg<(A+j)->avg) EX(A+i,A+j); } } }; void main() { int s=-1; printf("请输入学生个数 "); scanf("%d",&s); Student *all=new Student[s]; printf("eng,puter "); for(int i=0;i<s;i++) { scanf("%d %d %d",&(all+i)->eng,&(all+i)->math,&(all+i)-&puter);//输入以空格为间隔以回车为确定 (all+i)->Aveage();(all+i)->Sum();(all+i)->id=i+1; } PX(all,s); printf("ID,eng,puter,avg,total "); for(int j=0;j<s;j++) { printf("%d,%d,%d,%d,%2.2f,%d ",(all+j)->id,(all+j)->eng,(all+j)->math,(all+j)-&puter,(all+j)->avg,(all+j)->total); } }

c语言编程实例

#include<stdio.h> int main(void) { int i,j; for(i = 1;i <=4;i++) { for(j = 1;j <= i;j++) printf("%d",j); printf(" "); } printf(" "); return 0; } 这是最简单的，可以有附加功能。

求C语言程序设计实例，要100行以上

Problem Description: The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will , in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each .

For a given request list, you are pute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input: There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output Print the total time on a single line for each test case.

Sample Input 1 2 3 2 3 1 0

Sample Output 17 41

Here is my code:

#include<stdio.h> int time(int a,int b) { if(a==b)return 0; else if(a<b) return (b-a)*6; else return (a-b)*4; } int main() { int n,start,end,ans; while(1) { scanf("%d",&n); if(n==0)break; start=0;ans=0; while(n--) { scanf("%d",&end); ans+=time(start,end)+5; start=end; } printf("%d ",ans); } return 0; } 说明：程序代码的多少不是评价一个程序质量的唯一标准。