实验一
task1
#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; }
#include <stdio.h> int main() { printf(" O O\n"); printf("<H> <H>\n"); printf("I I I I\n"); return 0; }
task2
#include <stdio.h> int main() { int n,sum; scanf("%d",&n); sum = n*(n+1)/2; printf("sum = %d\n",sum); return 0; }
写法1、2、3都能实现题目要求,写法4运算时优先运算(n+1)和2*n两个部分,其次运算前者除后者,因此会出现错误
task4
#include <stdio.h> int main() { int x, t, m; x = 123; printf("x = %d\n", x); t = 0; m = x % 10; t = t * 10 + m; x = x / 10; m = x % 10; t = t * 10 + m; x = x / 10; m = x % 10; t = t * 10 + m; x = x / 10; printf("t = %d\n", t); return 0; }
(2)实现的功能是把个位与百位交换
task5
#include <stdio.h> int main() { float a, b, c; scanf("%f%f%f", &a, &b, &c); if(a+b>c,a+c>b,b+c>a) printf("能够成三角形\n"); else printf("不能够成三角形\n"); return 0; }
task6
#include <stdio.h> int main() { int year; int x; x = 1000000000; year = x/60/60/24/365; printf("10亿秒约等于%d年\n",year); return 0; }
task7
#include <stdio.h> #include <stdlib.h> #include <time.h> int main() { int n; srand(time(0)); n = rand() % 41 + 60; printf("n = %d\n", n); return 0; }
task8
#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' or ans1 == 'Y') && (ans2 == 'y' or ans2 == 'Y')) printf("罗马不是一天建成的,继续保持哦:)\n"); else printf("罗马不是一天毁灭的,我们来建设吧\n"); return 0; }