实验1
任务1:
task1_1
#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"); }
task1_2:
#include<stdio.h> int main() { printf(" O O \n"); printf("<H> <H>\n"); printf("I I I I\n"); }
任务2:
#include <stdio.h> int main() { double a, b, c scanf("%lf%lf%lf", &a, &b, &c); 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; }
任务3:
#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; }
回答问题:跳过了第二问的输入,这一行可以避免第二问直接识别到enter跳过回答
任务4:
#include<stdio.h> int main() { double x, y; char c1, c2, c3; int a1, a2, a3; scanf("%d,%d,%d", &a1, &a2, &a3);//缺少& 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);//双精度浮点数用%lf printf("x = %f, y = %lf\n",x, y); return 0; }
任务5:
#include <stdio.h> int main() { int year; year = 1000000000/(60*60*24*365); if(1000000000%(60*60*24*365)>60*60*24*365/2) year++; printf("10亿秒约等于%d年\n", year); return 0; }
任务6:
#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; }
任务7:
#include <stdio.h> #include <math.h> int main() { double c, f; while(scanf("%lf", &c) != EOF) { f = c*9/5+32; printf("摄氏度c=%.2f,华氏度f=%.2f\n", c, f); printf("\n"); } return 0; }
任务8:
#include <stdio.h> #include <math.h> int main() { double a,b,c,area,s,d; while(scanf("%lf", &a) != EOF) { scanf("%lf", &b); scanf("%lf", &c); s = (a+b+c)/2; d=s*(s-a)*(s-b)*(s-c); area = pow(d,0.5); printf("a = %lf,b = %lf,c = %lf,area = %.3f\n",a,b,c,area); } return 0;