实验二 

 1 #include <stdio.h>
 2 int main()
 3 {
 4     int a = 5, b = 7, c = 100, d, e, f;
 5 
 6     d = a / b * c;//0
 7     e = a * c / b;//71
 8     f = c / b * a;//70
 9     printf("%d,%d,%d", d, e, f);
10 
11     return 0;
12 }

 

 

 

 

 1 #include <stdio.h>
 2 int main()
 3 {
 4     int x = 1234;
 5     float f = 123.456;
 6     double m = 123.456;
 7     char ch = 'a';
 8     char a[] = "Hello World!";
 9     int  y = 3, z = 4;
10 
11     printf("%d %d\n",y,z);
12     printf("y=%d,z=%d\n",y,z);
13     printf("%8d,%2d\n",x,x);
14     printf("%f,%8f,%8.1f,%0.2f,%.2e\n",f,f,f,f,f);
15     printf("%lf\n",m);
16     printf("%3c\n",ch);
17     printf("%s\n%16s\n%10.5s\n%2.5s\n%.3s\n%.1s\n",a,a,a,a,a,a);
18 
19     return 0;
20 }

 

 

 

 1 #include <stdio.h>
 2 int main()
 3 {
 4     double x, y;
 5     char c1, c2, c3;
 6     int a1, a2, a3;
 7 
 8     scanf_s("%d%d%d", &a1, &a2, &a3);//此处未用'&'
 9     printf("%d,%d,%d\n", a1, a2, a3);
10     scanf_s("%c%c%c", &c1, &c2, &c3);
11     printf("%c%c%c\n", c1, c2, c3);
12     scanf_s("%lf,%lf", &x, &y);//double对应lf而不是f
13     printf("%lf,%lf\n", x, y);
14 
15     return 0;
16 }

 

 

 

 

 1 #include <stdio.h>
 2 int main()
 3 {
 4     char a;
 5     a = getchar();
 6     if (a>='A'&&a<='Z'||a>='a'&&a<='z')
 7         printf("%c为英文字母", a);
 8     else if (a >= '0' && a <= '9')
 9         printf("%c为数字", a);
10     else
11         printf("%c为其它字符",a);
12     
13     return 0;
14 }

 

 

 

 

 

 

 

 1 #include <stdio.h>
 2 int main() {
 3     char ans1, ans2;
 4 
 5     printf("复习了没? (输入y或Y表示复习了，输入n或N表示没复习) :  ");
 6     ans1 = getchar();  // 从键盘输入一个字符，赋值给ans1
 7 
 8     getchar();// 思考这里为什么要加这一行 
 9 
10     printf("\n动手敲代码了没? (输入y或Y表示敲了，输入n或N表示木有敲) :  ");
11     ans2 = getchar();
12 
13     if ((ans1=='y'||ans1=='Y')&&(ans2=='y'||ans2=='Y'))
14         printf("\n罗马不是一天建成的:)\n");
15     else
16         printf("\n罗马不是一天毁灭的。。。\n");
17 
18     return 0;
19 }

 

 

 

 

 

 

 1 #include <stdio.h>
 2 #include <math.h>
 3 int main()
 4 {
 5     int n,sum;
 6     scanf_s("%d", &n);
 7     sum = (1 - pow(2, n+1)) / (1 - 2);
 8     printf("%d", sum);
 9 
10     return 0;
11 }

 

 

posted @ 2020-11-06 08:11  违规昵称4E32  阅读(135)  评论(0)    收藏  举报