谭浩强版本C语言课后习题第四章

Posted 2022-05-04 1998wdq

2、对于逻辑量表达式1表示真，0表示假。

     对于逻辑量表达式非0表示真，0表示假。

3、值得注意的是c语言中的优先级，优先级由高到低依次为：

4、

 1 #include<stdio.h>
 2 int main()
 3 
 4     int a,b,c,t;
 5     scanf("%d,%d,%d",&a,&b,&c);
 6     if(a<b)
 7     t=b;
 8      a=b;
 9      b=t;
10     if(a<c)
11     t=c;
12     a=c;
13     c=t;
14     printf("最大的是：%d\\n",a);
15     return 0;
16 

法2：

 1 #include<stdio.h>
 2 int main()
 3 
 4     int a,b,c,temp,max;
 5     scanf("%d,%d,%d",&a,&b,&c);
 6     temp=(a>b)?a:b;
 7     max=(temp>c)?temp:c;
 8     printf("最大的是：%d\\n",max);
 9     return 0;
10 

5、

 1 #include<stdio.h>
 2 #include<math.h>
 3 int main()
 4 
 5     int a,b;
 6     scanf("%d",&a);
 7     if(a>=1000)
 8     printf("enter again:\\n");
 9     scanf("%d",&a);
10     b=sqrt(a);
11     printf("%d\\n",b);
12     return 0;
13 

8、

 1 #include<stdio.h>
 2 int main()
 3 
 4     float score;
 5     char grad;
 6     printf("enter score:");
 7     scanf("%f",&score);
 8     while(score>100||score<0)
 9     printf("enter again:");
10     scanf("%f",&score);
11     
12     switch((int)(score/10))
13     case 10:
14     case 9:grad=‘A‘;break;
15      case 8:grad=‘B‘;break;
16      case 7:grad=‘C‘;break;
17       case 6:grad=‘B‘;break;
18      case 5:
19      case 4:
20      case 3:
21      case 2:
22      case 1:
23      case 0:grad=‘E‘;
24     
25     printf("成绩是：%5.1f,等级是： %c\\n",score,grad);
26     return 0;
27 

 9、给一个不多于五位的正整数，奥球

求出它是几位数，分别是出每一位数字，按逆序输出各位数字

 1 #include<stdio.h>
 2 #include<math.h>
 3 int main()
 4 
 5     int num,indiv,ten,hundred,thousand,ten_thousand,place;
 6     printf("请输入一个不多于5位的整数:");
 7     scanf("%d",&num);
 8     if(num>9999)
 9         place=5;
10     else
11         if(num>999)
12             place=4;
13         else
14             if(num>99)
15                 place=3;
16             else
17                 if(num>9)
18                     place=2;
19                 else
20                     place=1;
21                 printf("位数：%d\\n",place);
22                 printf("每位数字为\\n");
23                 ten_thousand=num/10000;
24                 thousand=(int)(num-ten_thousand*10000)/1000;
25                 hundred=(int)(num-ten_thousand*10000-thousand*1000)/100;
26                 ten=(int)(num-ten_thousand*10000-thousand*1000-hundred*100)/10;
27                 indiv=(int)(num-ten_thousand*10000-thousand*1000-hundred*100-ten*10);
28                 switch(place)
29                 
30 case 5:printf("%d,%d,%d,%d,%d",ten_thousand,thousand,hundred,ten,indiv);
31       printf("\\n反序数字为\\n");
32       printf("%d%d%d%d%d",indiv,ten,hundred,thousand,ten_thousand);
33       break;
34 case 4:printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
35       printf("\\n反序数字为\\n");
36       printf("%d%d%d%d",indiv,ten,hundred,thousand);
37       break;
38 case 3:printf("%d,%d,%d",hundred,ten,indiv);
39       printf("\\n反序数字为\\n");
40       printf("%d%d%d",indiv,ten,hundred);
41       break;
42 case 2:printf("%d,%d",ten,indiv);
43       printf("\\n反序数字为\\n");
44       printf("%d%d",indiv,ten);
45       break;
46 case 1:printf("%d",indiv);
47       printf("\\n反序数字为\\n");
48       printf("%d",indiv);
49       break;
50                 
51                 return 0;
52 

12、判断是否在塔外

 1 #include<stdio.h>
 2 int main()
 3 
 4     int h=10;
 5     float x1=2,y1=2,x2=-2,y2=2,x3=-2,y3=-2,x4=2,y4=-2,x,y,d1,d2,d3,d4;
 6     printf("请输入x,y: ");
 7     scanf("%f,%f",&x,&y);
 8     d1=(x-x4)*(x-x4)+(y-y4);
 9     d2=(x-x1)*(x-x1)+(y-y1)*(y-y1);
10     d3=(x-x2)*(x-x2)+(y-y2)*(y-y2);
11     d4=(x-x3)*(x-x3)+(y-y3)*(y-y3);
12     if(d1>1&&d2>1&&d3>1&&d4>1)
13         h=0;
14     printf("高度为:%d\\n",h);
15     return 0;
16