实验1

1.1

include <stdio.h>

int main() {
printf(" O \n");
printf("\n");
printf("I I\n");
printf(" O \n");
printf("\n");
printf("I I\n");
return 0;
}

1

1.2

include <stdio.h>

int main() {
printf(" O ");
printf("\t");
printf(" O \n");
printf("");
printf("\t");
printf("\n");
printf("I I");
printf("\t");
printf("I I\n");
return 0;
}

2

2.0

include <stdio.h>

int main(){
// 从键盘上输入三个数据作为三角形边长,判断其能否构成三角形
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
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;
}

3

3.0

include <stdio.h>

int main() {
char ans1, ans2;// 用于保存用户输入的答案
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(); // 从键盘输入一个字符,赋值给ans1

char c = getchar(); // 思考这里为什么要加这一行
//printf("%d",c);结果是10,所以是用来吞掉换行符的
    
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;
}

3

4.0

include<stdio.h>

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

scanf("%d%d%d", &a1, &a2, &a3);//标注错误:未加&
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);//标注错误:double对应的是ld,而且没有逗号
printf("x = %lf, y = %lf\n",x, y);
return 0;
}

4

5.0

include<stdio.h>

int main()
{
int year;
year = (1e9)/(606024*365.0) + 0.5;
printf("10亿秒约等于%d年\n", year);
return 0;
}

5

6.0

include <stdio.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");
}
return 0;
}

6

7.0

include <stdio.h>

int main(){
double x,ans;
while(scanf("%lf", &x) != EOF)
{
ans = (9.0/5)*x + 32;
printf("摄氏度为%.2f时,华氏温度为%.2f\n", x, ans);
printf("\n");
}
return 0;
}

7

8.0

include <stdio.h>

include <math.h>

int main() {
int a,b,c;
double p,area;
while (scanf("%d%d%d",&a,&b,&c) != EOF) {
p = (a+b+c)/2.0;
area = sqrt(p(p-a)(p-b)*(p-c));
printf("a = %d,b = %d,c = %d, area = %.3lf\n\n",a,b,c,area);
}
return 0;

8

posted @ 2025-10-08 17:50  135hlj  阅读(8)  评论(1)    收藏  举报