谭浩强C语言程序设计第四版例题与课后题答案

1、第四章选择程序结构设计4.1选择结构和条件判断例子4.1求二元一次方程的根#include<stdio.h>#include<math.h>int main()double a,b,c,disc,x1,x2,p,q;scanf("%lf%lf%lf",&a,&b,&c);/double &a表示a在内存中的地址disc=b*b-4*a*c;if(disc<0)printf("This equation hasnt real rootsn");elsep=-b/(2.0*a);q=sqrt(di

2、sc)/(2.0*a);x1=p+q;x2=p-q;printf("x1=%7.2fnx2=%7.2fn",x1,x2);/指定数据占7列小数点占2列return 0;4.2用if语句实现选择结构#include<stdio.h>int main()int a,b;printf("please input a,b:");scanf("%d%d",&a,&b); /输入的形式应该注意if(a>b)printf("max=%d",a);elseprintf("max=%d&q

3、uot;,b);return 0;/ 输入两个数判断较大者#include<stdio.h>int main()int a,b;printf("please input a,b:");scanf("%d%d",&a,&b); /输入的形式应该注意if(a>b)printf("max=%d",a);elseprintf("max=%d",b);return 0;*/输入两个数由小到大排列#include<stdio.h>int main()int a,b,t;printf

4、("please input a,b:");scanf("%d%d",&a,&b); /输入的形式应该注意1,2if(a>b)t=a;a=b;b=t;printf("%d%d",a,b);elseprintf("%d%d",a,b);return 0;/*#include<stdio.h>int main()int a,b,t;printf("please input a,b:");scanf("%d%d",&a,&b); /

5、输入的形式应该注意1,2if(a>b)t=a;a=b;b=t;printf("%d%d",a,b);return 0;*/4.3输入3个数比较大小a>b交换aba>c比较b>c比较思想运算符的优先级高->低算术运算符关系运算符赋值运算符多看一些例子：如c>a+b <=> c>(a+b)c>b=c <=> (c>b)=ca=b<c <=> a=(b<c)a=b>c <=> a=(b>c)4.3.2关系表达式真假4.4逻辑运算符和逻辑表达式优先级（高-&

6、gt;低）！（非）算术运算符关系运算符&&和|赋值运算符例如：（a>b）&&（x>y） <=> a>b && x>y（a=b）|（x=y） <=> a=b|x=y注意：关系运算符<<=>>=以上（以下）的优先级相同但是上面大于下面的=！=4.5条件运算符和条件表达式if(a>b)max=a;elsemax=b;<=>max=(a>b)?a:b;例4.4/输入字符如果大写转换小写，小写不变#include<stdio.h>int main()

7、char ch;scanf("%c",&ch);ch=(ch>A&&ch<=Z)?(ch+32):ch;printf("%cn",ch);return 0;4.6选择结构的嵌套4.7用switch语句实现多分支选择结构注意：关键字break别写错了，记得考试的时候就忘记了。#include<stdio.h>int main()void action1(int,int),action2(int,int);char ch;int a=15,b=23;ch=getchar();switch(ch)casea:ac

8、tion1(a,b);break;/caseA:action1(a,b);break;caseA:action1(a,b);break;caseb:/caseB:action2(a,b);break;caseB:action2(a,b);break;default:putchar(a);return 0;void action1(int x,int y)printf("x+y=%dn",x+y);void action2(int x,int y)printf("x*y=%dn",x*y);4.8选择结构程序综合举例例子4.8判断闰年#include<

9、;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;elseleap=0;if(leap)printf("闰年");elseprintf("平年");return 0;4.9求 a*x*x+b*x+c=0方程的解（1）a=0不是二元一次方程（2）b*b-4*a*c=0有两个相等的实根（3）b*b-4*a*

10、c>0有两个不等的实根（4）b*b-4*a*c<0有两个共轭复根。应当以p+q*i和p-q*i的形式输出复根。其中p=-b/2a,q=(sqrt(b*b-4*a*c)/(2a)#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)prin

11、tf("is not a quadraticn");elsedisc=b*b-4*a*c;if(fabs(disc)<=1e-6)printf("has two equal roots:%8.4fn",-b/(2*a);elseif(disc>1e-6)x1=(-b+sqrt(disc)/2*a;x2=(-b-sqrt(disc)/2*a;printf("has distinct real roots:%8.4f and %8.4fn",x1,x2);elserealpart=-b/(2*a);imagpart=sqrt(-disc)/(2*a);printf("has complex roots:n");printf("%8.4f+%8.4fin",realpart,imagpart);printf("%8.4f-%8.4fin",realpart,i