实验一

task1_1.c

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 = %lf, y = %lf\n",x, y);
    
    return 0;
}

运行结果

 task1_2.c

源代码

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

运行结果

task2.c

源代码

#include <stdio.h>
int main()
{
    float a, b, c;
    // 输入三边边长
    scanf("%f%f%f", &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");
    system("pause");
    return 0;
}

运行结果

 task3.c

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 ((ans2 == 'y' && ans1 == 'y') ||
        (ans1 == 'Y' && ans2 == 'Y') ||
        (ans1 == 'Y' && ans2 == 'y') ||
        (ans1 == '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");
    system("pause");
    return 0;
}

运行结果

task4.c

源代码

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 int main()
 4 {
 5     double x, y;
 6     char c1, c2, c3;
 7     int a1, a2, a3;
 8 
 9     scanf("%d %d %d", &a1, &a2, &a3);
10     printf("a1 = %d, a2 = %d, a3 = %d\n", a1, a2, a3);
11     getchar();
12     scanf("%c%c%c", &c1, &c2, &c3);
13     printf("c1 = %c, c2 = %c, c3 = %c\n", c1, c2, c3);
14 
15     scanf("%lf%lf", &x, &y);
16     printf("x = %lf, y = %lf\n", x, y);
17 
18     system("pause");
19     return 0;
20 }

运行结果

 task5.c

源代码

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

    const double second = 1e9;
    double temp = ((second / 3600) / 24) / 365;
    if (temp - (int)temp < 0.5) {
        year = (int)temp;
    }
    else {
        year = (int)temp + 1;
    }

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

运行结果

 task6.c

源代码

 1 #include <stdio.h>
 2 #include <math.h>
 3 int main()
 4 {
 5     double x, ans;
 6     while (scanf("%lf", &x) != EOF) {
 7         ans = pow(x, 365);
 8         printf("%.2f的365次方: %.2f\n", x, ans);
 9         printf("\n");
10     }
11 
12     return 0;
13 }

运行结果

 task7.c

源代码

#include <stdio.h>

int main()
{
    double f;
    double c;

    while (scanf("%lf", &c) != EOF) {
        f = 9.0 / 5.0 * c + 32.0;
        printf("摄氏度c = %.2lf时, 华氏度f = %.2lf\n", c, f);
        printf("\n");
    }

    return 0;
}

运行结果

 task8.c

源代码

#include <stdio.h>
#include <math.h>

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

    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 = %d, b = %d, c = %d, area = %.3lf\n", (int)a, (int)b, (int)c, area);
        printf("\n");
    }

    return 0;
}

运行结果