实验1
task1-1.c
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
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"); return 0; }
运行结果
task2
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"); return 0; }
运行结果
task3
源代码
#include <stdio.h> int main3() { 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"); return 0; }
运行结果
task4
源代码
#include<stdio.h> int main4() { 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; }
运营结果
task5
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); return 0; }
运营结果
task6
源代码
#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; }
运营结果
task7
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; }
运营结果
task8
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; }
运营结果