task1
实验1
1 #include <stdio.h> 2 int main() 3 { 4 printf(" O \n"); 5 printf("<H>\n"); 6 printf("I I\n"); 7 printf(" O \n"); 8 printf("<H>\n"); 9 printf("I I\n"); 10 return 0; 11 }

1 #include <stdio.h> 2 int main() 3 { 4 printf(" O O\n"); 5 printf("<H> <H>\n"); 6 printf("I I I I\n"); 7 return 0; 8 }

实验二
1 #include <stdio.h> 2 int main() 3 { 4 double a, b, c; 5 scanf_s("%lf%lf%lf", &a, &b, &c); 6 if ((a + b > c) && (a + c > b) && (b + c > a)) 7 printf("能构成三角形\n"); 8 else 9 printf("不能构成三角形\n"); 10 return 0; 11 }



实验三
1 #include <stdio.h> 2 int main() 3 { 4 char ans1, ans2; 5 6 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) : "); 7 ans1 = getchar(); 8 9 getchar(); 10 11 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) : "); 12 ans2 = getchar(); 13 14 if ((ans1=='y'|| ans1 == 'Y') && (ans2 == 'y' || ans2 == 'Y')) 15 printf("\n罗马不是一天建成的, 继续保持哦:)\n"); 16 else 17 printf("\n罗马不是一天毁灭的, 我们来建设吧\n"); 18 19 return 0; 20 }


如果删除line9,如图,直接跳过ans2的赋值,因为/n被getchar()读取

实验四
1 #include<stdio.h> 2 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 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 = %f, y = %lf\n", x, y); 17 18 return 0; 19 }

实验五
1 #include <stdio.h> 2 3 int main() 4 { 5 int year; 6 7 year = 1000000000 / (365 * 24 * 60 * 60); 8 9 printf("10亿秒约等于%d年\n", year); 10 return 0; 11 }

实验六
1 #include <stdio.h> 2 #include <math.h> 3 4 int main() 5 { 6 double x, ans; 7 8 while (scanf("%lf", &x) != EOF) 9 { 10 ans = pow(x, 365); 11 printf("%.2f的365次方: %.2f\n", x, ans); 12 printf("\n"); 13 } 14 return 0; 15 }

实验七
1 #include <stdio.h> 2 3 int main() { 4 double c, f; 5 while (scanf("%lf", &c) != EOF) { 6 f = 9.0 / 5.0 * c + 32; 7 printf("摄氏度c = %.2f时,华氏度f = %.2f\n", c, f); 8 } 9 return 0; 10 }

实验八
1 #include <stdio.h> 2 #include <math.h> 3 int main() { 4 double a, b, c, s, area; 5 while (scanf("%lf%lf%lf", &a, &b, &c) != EOF) { 6 s = (a + b + c) / 2.0; 7 area = sqrt(s * (s - a) * (s - b) * (s - c)); 8 printf("a=%.0f, b=%.0f, c=%.0f, area=%.3f\n", a, b, c, area); 9 } 10 return 0; 11 }

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