第一次实验作业

实验任务一

源代码:

#include<stdio.h>
int main()
{
	printf(" O \n");
	printf("<H>\n");
	printf("I I\n");
	printf(" O \n");
	printf("<H>\n");
	printf("I I\n");
	
	
	return 0;
}

  

 



运行结果:

80c955a1726305250a6f2bf03d50267b_3779923-20260323164004030-1056928155

 

 源代码(水平):

#include<stdio.h>
#include<stdlib.h>
int main()
{

    printf(" o     o\n");
    printf("<H>   <H>\n");
    printf("I I   I I\n");

    system("pause");
return 0;
}

  

运行结果:

850037269d8290e70916fa2d9c80b2fe_3779923-20260323163400812-990898505

 

实验任务2

源代码:

 
#include <stdio.h>

int main()
{
    double a, b, c;

    // 输入三边边长
    scanf("%lf%lf%lf", &a, &b, &c);

    // 判断能否构成三角形
    // 补足括号里的逻辑表达式
    if((a+b>c)&&(a+c>b)&&(b+c>a))
        printf("能构成三角形\n");
    else
        printf("不能构成三角形\n");

    return 0;
}

  

运行结果:

de8057baf461e27b8c1d050ba05d5a7a_3779923-20260323164914104-439120354

1e3b526be680403a538bdbda9ce83381_3779923-20260323165022376-1459248834

 

实验任务3

源代码:

 
#include <stdio.h>
int main()
{
    char ans1, ans2;

    printf("每次课前认真预习、课后及时复习了没? (输入y或Y表示有,输入n或N表示没有) :  ");
    ans1 = getchar();

    getchar();

    printf("\n动手敲代码实践了没? (输入y或Y表示敲了,输入n或N表示木有敲) :  ");
    ans2 = getchar();

    if ((ans1=='y'||ans1=='Y')&&(ans2=='y'||ans2=='Y'))
        printf("\n罗马不是一天建成的, 继续保持哦:)\n");
    else
        printf("\n罗马不是一天毁灭的, 我们来建设吧\n");

    return 0;
}

  

运行结果:

1d8b39a06a34a1ba310897b49974deb0_3779923-20260323165452637-1681353560

 

实验任务4

源代码:

 

#include<stdio.h>

int main()
{
    double x, y;
    char c1, c2, c3;
    int a1, a2, a3;

    scanf("%d%d%d", &a1, &a2, &a3);
    printf("a1 = %d, a2 = %d, a3 = %d\n", a1,a2,a3);

    scanf("%c%c%c", &c1, &c2, &c3);
    printf("c1 = %c, c2 = %c, c3 = %c\n", c1, c2, c3);

    scanf("%lf,%lf", &x, &y);
    printf("x = %f, y = %lf\n",x, y);

    return 0;
}

  

运行结果:

fbea179f9038a244c9ea6d95817f1e72_3779923-20260323170644169-31315646

 

 

实验任务5

源代码:

 
#include <stdio.h>

int main()
{
    int year;

    double seconds = 1e9;
    double seconds_per_year = 365.0 * 24 * 3600;
    double years = seconds / seconds_per_year;

    year = (int)(years + 0.5);

    printf("10亿秒约等于%d年\n", year);
    return 0;
}

  

运行结果:

a4601bce0ac0b36162b1ecd865cf5643_3779923-20260323170815820-844078286

 

实验任务6

源代码:

 
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
{
    double x, ans;

    while(scanf("%lf",&x)!= EOF)
    {
        ans = pow(x, 365);
        printf("%.2f的365次方:%.2f\n", x, ans);
        printf("\n");
    }
    system("pause");
    return 0;
}

  

 

运行结果:

cea81bddfb60875eed411df86fd81c0b_3779923-20260323171045944-504179664

 

实验任务7

源代码:

 
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
{
    double F,C;

    while(scanf("%lf", &C)!= EOF)
    {
        F = 9.0/5.0*C + 32;
        printf("%.2f摄氏度转化为%.2f华氏度\n",C, F);
        printf("\n");
    }
    system("pause");
    return 0;
}

  

运行结果:

 

81a9f38f1413844350775ec9ceb7329c_3779923-20260323171249883-1358500748

 

实验任务8

源代码:

 
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
{
    double area, a, b, c, s;
    while(scanf("%lf%lf%lf", &a, &b, &c) != EOF)
    {
        s = (a+b+c)/2;
        area = sqrt(s*(s-a)*(s-b)*(s-c));
        printf("a = %.3f, b = %.3f, c = %.3f, area = %.3f\n",a, b, c, area);
        printf("\n");
    }
    system("pause");
    return 0;
}

  

运行结果:

63821363e624c51a383fc11e8a6a73f7_3779923-20260323171413582-576998145

 

 
 
 
posted @ 2026-03-24 23:52  悲绪  阅读(9)  评论(0)    收藏  举报