第四章选择程序设计

例4.1 解二元一次方程

 #include<stdio.h>
 #include<math.h>
 int main()
 {
  double a,b,c,disc,x1,x2,p,q;
  scanf("%lf%lf%lf",&a,&b,&c);
  disc=b*b-4*a*c;
  if(disc<0)
 printf("This eaquation hasn't real roots\n");
  else 
 { p=-b/(2.0*a);
 q=sqrt(disc)/(2.0*a);
 x1=p+q;x2=p-q;
 printf("real roots:\nx1=%7.2f\nx2=%7.2f\n",x1,x2);
 }
 return 0;
 }

运行结果如下

例4.2 输入两个实数 按由小到大的顺序输出这两个数

#include<stdio.h>
int main()
float a,b,t;
scanf("%f,%f",&a,&b);
if(a>b)
{
t=a;
a=b;
b=t;
}
printf("%5.2f,%5.2f\n",a,b);
return 0;
}

运行结果如下

例4.3 输入三个数a,b,c，要求按照由小到大顺序输出

#include<stdio.h>
int main()
{
float a,b,c,t;
scanf("%f,%f,%f",&a,&b,&c);
if(a>b)
{
t=a;
a=b;
b=t;
}
if(a>c)
{
 t=a;
a=c;
c=t;
}
if(b>c)
{
 t=b;
b=c;
c=t;
}
 printf("%5.2f,%5.2f,%5.2f\n",a,b,c);
 return 0;
}

运行结果如下

例4.4 输入一个字符，判断它是否为大写字母，如果是，将它转换为小写字母，如果不是，不转换，然后输出最后得到的字符

#include<stdio.h>
int main()
{
 char ch;
scanf("%c",&ch);
ch=(ch>='A'&&ch<='Z')?(ch+32):ch;
printf("%c\n",ch);
 return 0;
}

运行结果如下

例4.5 有一阶跃函数 输入X值 要求输出相应Y值

#include<stdio.h>
int main()
{
 int x,y;
scanf("%d",&x);
if(x<0)
 y=-1;
else 
 if(x==0)y=0;
 else y=1;
 printf("x=%d,y=%d\n",x,y);
 return 0;
}

#include<stdio.h>
int main()
{
int x,y;
scanf("%d",&x);
if(x>=0)
 if(x>0) y=1;
 else y=0;
 else y=-1;
 printf("x=%d,y=%d\n",x,y);
 return 0;
}

运行结果如下

例4.6 要求按照考试成绩的等级输出百分制分数段，成绩的等级由键盘输入

#include<stdio.h>
int main()
{
char grade;
scanf("%c",&grade);
printf("Your score:");
switch(grade)
{
case'A':printf("85~100\n");break;
case'B':printf("70~84\n");break;
case'C':printf("60~69\n");break;
case'D':printf("<60\n");break;
default:printf("enter data error!\n");
}
 return 0;
}

运行结果如下

例4.7 用switch语句处理菜单命令

#include<stdio.h>
int main()
{
void action1(int,int),action2(int,int);
char ch;
int a=15,b=23;
ch=getchar();
switch(ch)
{
case'a':
case'A':action1(a,b);break;
case'b':
case'B':action2(a,b);break;
default:putchar('\a');
}
return 0;
}
void action1(int x,int y)
{
 printf("x+y=%d\n",x+y);
}
void action2(int x,int y)
{
 printf("x*y=%d\n",x*y);
}

运行结果如下

例4.8

1 写一程序，判断某一年是否为闰年。

#include<stdio.h>
int main()
{
int year,leap;
printf("enter year:");
scanf("%d",&year);
if(year%4==0)
{
 if(year%100==0)
 {
  if(year%400==0)
  leap=1;
 else leap=0;
}
else 
 leap=1;
}
else
 leap=0;
if(leap)
 printf("%d is ",year);
else 
 printf("%d is not ",year);
printf("a leap year.\n");
return 0;
}

运行结果如下

2 将程序中第7-12行改写成以下的if语句

#include<stdio.h>
int main()
{
int year,leap;
printf("enter year:");
scanf("%d",&year);
if(year%4!=0)
 leap=0;
else if(year%100!=0)
 leap=1;
else if(year%400!=0)
 leap=0;
else
 leap=1;
if(leap)
 printf("%d is ",year);
else 
 printf("%d is not ",year);
printf("a leap year.\n");
return 0;
}

运行结果如下

3 将上述if语句用下面的if语句代替

#include<stdio.h>
int main()
{
int year,leap;
printf("enter year:");
scanf("%d",&year);
if((year%4==0&&year%100!=0)||(year%400==0))
 leap=1;
else
 leap=0;
if(leap)
 printf("%d is ",year);
else 
 printf("%d is not ",year);
printf("a leap year.\n");
return 0;
}

运行结果如下

例4.9 求二元一次方程解

#include<stdio.h> 
#include<math.h>
int main()
{
double a,b,c,disc,x1,x2,realpart,imagpart;
scanf("%lf,%lf,%lf",&a,&b,&c);
printf("The equation");
if(fabs(a)<=1e-6)
 printf("is not a quadratic\n");
else
{
 disc=b*b-4*a*c;
 if(fabs(disc)<=1e-6)
  printf("has two equal roots:%8.4f\n",-b/(2*a));
 else
  if(disc>1e-6)
  {
   x1=(-b+sqrt(disc))/(2*a);
   x2=(-b-sqrt(disc))/(2*a);
   printf("has distinct real roots:%8.4f and %8.4f\n",x1,x2);
  }
  else
  {
   realpart=-b/(2*a);
   imagpart=sqrt(-disc)/(2*a);
   printf("has complex roots:\n");
   printf("%8.4f+%8.4fi\n",realpart,imagpart);
   printf("%8.4f-%8.4fi\n",realpart,imagpart);
  }
}
return 0;
}

运行结果如下

#include<stdio.h>
#include<math.h>
int main()
{
int c,s;
float p,w,d,f;
printf("please enter price,weight,discount:");
scanf("%f,%f,%d",&p,&w,&s);
if(s>=3000) c=12;
else c=s/250;
switch(c)
{
case 0:d=0;break;
case 1:d=2;break;
case 2:
case 3:d=5;break;
case 4:
case 5:
case 6:
case 7:d=8;break;
case 8:
case 9:
case 10:
case 11:d=10;break;
case 12:d=15;break;
}
f=p*w*s*(1-d/100);
printf("freight=%10.2f\n",f);
return 0;
}

运行结果如下