第四章 选择结构程序设计

第四章

例4.1

#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 equation 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 root:\nx1=%7.2f\nx2=%7.2f\n",x1,x2);
}
return 0;
}

结果为：

4.2 输入两个实数，按由大到小排序

#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
}

结果为：

4.3

#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;
}

结果为：
4.4

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

结果为：

4.5

#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;
}

结果为：

4.6

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

结果为：

4.7

#include<stdio.h>
int main()
{void actino1(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);
}

结果为：
4.8

#include<stdio.h>
int main()
{
int year,leap;
printf("enter year:");
scanf("%d",&year);
if(year%4==0)
{
if(year%100==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;
}

结果为：

4.9

#include<stdio.h>
#include<math.h>
int mian()
{
double a,b,c,disc,x1,x2,realpart,imagpart;
scanf("%lf,%lf,%lf",&a,&b,&c);
printf("The equation");
if(fabs(a)<=le-6)
printf("is not aquadratic\n");
else
{
disc=b*b-4*4*c;
if(fabs(disc)<=le-6)
printf("has two equal root:%8.4f\n",-b/(2*a));
else
if(disc>le-6)
}
x1=(-b+sqrt(disc))/(2*a);
x2=()-b-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;
}

结果为：