C language design lab report (second)

---

C Programming lab report

• Name: Kung

• Experimental Location: home

• Experiment time: 2019.3.9

• Experimental project: a character code and ASCLL ; application operator of the expression ; the sequence structure of the application ; mathematical function ; chickens and rabbits with cage ; determining the coordinates .

---

1, experimental purposes and requirements

1. Master ++ increment, decrement - operators used correctly.

2. Master basic input / output functions of the basic functions of the calling method, input / output control of a predetermined format and the like.

3. Master the method described algorithm structured program.

4. C language proficiency of the operator, i.e., operator precedence and tuberculosis, operation rule, type of operands.

Second, the experimental content

---

1、实验练习：Character code and ASCLL

1. brief description of the problem: Input Output Input Output characters written with different values

2. Experimental Code:

#include<stdio.h>
int main()
{
    //第一种写法 
    char ch;
    printf("请输入一个大写字母"); 
    ch=getchar();
    printf("小写字母为\n") ;
    putchar(ch+31);
    putchar(ch+32);
    putchar(ch+33);
    
    //第二种写法 
    scanf("%c",&ch);
    printf("%c,%c,%c\n",ch+31,ch+32,ch+33);
    
    //程序改进：计算平均值并输出 
    printf("%c",(ch+31+ch+32+ch+33)/3) ;
}

3. Analysis: input the character can be employed sacnf("%c",&字符变量)+ printf("%c",&字符变量）written; and may also be used ch=getc(stdin)``ch=getch()``ch=getchar()input, putc()``putch()``putchar()output.

---

2、实验练习：Application of the operator of the expression

1. brief description of the problem: the original data is broken

2. Experimental Code:

#include<stdio.h>
int main()
{
    int i,j,k;
    printf("请输入一个三位数");
    scanf("%d",&i) ;
    //第三位 
    k=i%10;
    //第二位
    j=(i%100-k)/10;
    //第一位 
    i/=100;
    printf("%d %d %d",k,j,i);
    
    return 0;
}

3. Analysis: At the time of data processing, and sometimes destroy data, then the data will be used next to the problem;

Solution 1: definition of a variable to store the data, and then destroyed when the variable retrieval;

Solution 2: Use the first final destruction.

---

3、实验练习：Calculation of medical expenses

1. brief description of the problem: None

2. Experimental Code:

#include<stdio.h>
int main()
{
    float i,j,k,x,y,z; 
    printf("请输入西药费，检查费，材料费，床位费，观察费，护理费：\n");
    scanf("%f,%f,%f,%f,%f,%f",&i,&j,&k,&x,&y,&z);
    float sum;
    sum=i+k+j+x+y+z;
    printf("应付款（元）：%.2f病人付款（元）：",sum);
    int pay;
    scanf("%d",&pay) ;
    printf("计算结果:\n病人付款=%d元\n应收款=%.2f元，应找回%.2f元",pay,sum,pay-sum);
    return 0;
}

3. Analysis: None

---

4、实验练习：Mathematical Functions

1. brief description of the problem: the variable digital hybrid computing expression

2. Experimental Code:

#include<stdio.h>
int main()
{
    float x;
    float y; 
    printf("请输入x的值：") ; 
    scanf("%f",&x);
    if(x<1){
        y=x;
    }else if(x<10){
        y=2*x-1;
    }else{
        y=3*x-1;
    }
    printf("y的值为%2f",y);
    return 0;
}

flow chart

3. Analysis: In the calculation of y=2x-1the input of the code when it should be y=2*x-1, *can not miss!

---

5、实验练习：Chickens and rabbits with cage

1. brief description of the problem: the same problem and fourth experiments

2. Experimental Code:

#include<stdio.h>
int main()
{
    int h,f,x,y;
    printf("鸡兔总数，鸡兔脚总数：") ;
    scanf("%d %d",&h,&f) ;
    if(x>0&&y>0) {
        x=(4*h-f)/2;
        y=(f-2*h)/2;
        printf("鸡有%d 兔有%d\n",x,y) ;
    }else{
        printf("输入错误！\n");
    }
    return 0;
}

flow chart

3. Analysis: The same experiment 4.

---

6、实验练习：Determining the coordinates

1. brief description of the problem: the expression of distance

2. Experimental Code:

#include<stdio.h>
int main()
{
    float x,y;
    printf("请输入一个点（x，y）：");
    scanf("%f,%f",&x,&y);
    if(x<0) {
        x=-x;
    }
    if(y<0) {
        y=-y;
    }
    if(((x-2)*(x-2)+(y-2)*(y-2))<1){
        printf("10");
    }else{
        printf("0");
    }
    
    return 0;
}

flow chart

3. Analysis: This is the most complex experiments to determine if the condition, at the beginning I use to calculate the distance from the origin to the center, to be used #include<math.h>in the sqrtrepresentation root. But there are drawbacks, so use the second method: Write the circle equation, <Ris the point within the circle. But pay attention to all the negative to positive.

---

Third, test summary

The experiment which is mainly to consolidate the basic C language written, some of the minutiae of error deepened the impression.

At the same time also studied writing markdown of (this post is written by markdown):

title:

# 一级标题
## 二级标题
### 三级标题

style:

*斜体* 

**粗体** 

==高亮==

~~删除线~~

and many more

back to the top