C++程序设计第三版(谭浩强)第四章习题答案

4.1 题
#include
using namespace std;
int main()
{int hcf(int,int);
int lcd(int,int,int);
int u,v,h,l;
cin>>u>>v;
h=hcf(u,v);
cout<<"H.C.F="<l=lcd(u,v,h);
cout<<"L.C.D="<return 0;
}
int hcf(int u,int v)
{int t,r;
if (v>u)
{t=u;u=v;v=t;}
while ((r=u%v)!=0)
{u=v;
v=r;}
return(v);
}
int lcd(int u,int v,int h)
{return(u*v/h);
}

4.2 题
#include
#include
using namespace std;
float x1,x2,disc,p,q;
int main()
{void greater_than_zero(float,float);
void equal_to_zero(float,float);
void smaller_than_zero(float,float);
float a,b,c;
cout<<"input a,b,c:";
cin>>a>>b>>c;
disc=b*b-4*a*c;
cout<<"root:"<if (disc>0)
{
greater_than_zero(a,b);
cout<<"x1="<}
else if (disc==0)
{equal_to_zero(a,b);
cout<<"x1="<}
else
{smaller_than_zero(a,b);
cout<<"x1="<cout<<"x2="<}
return 0;
}
void greater_than_zero(float a,float b) /* 定义一个函数，
用来求 disc>0 时方
程的根 */
{x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
}
void equal_to_zero(float a,float b) /* 定义一个函数，用
来求 disc=0 时方程
的根 */
{
x1=x2=(-b)/(2*a);
}
void smaller_than_zero(float a,float b) /* 定义一个函数，
用来求 disc<0 时方
程的根 */
{
p=-b/(2*a);
q=sqrt(-disc)/(2*a);
}

4.3 题
#include
using namespace std;
int main()
{int prime(int); /* 函数原型声明 */
int n;
cout<<"input an integer:";
cin>>n;
if (prime(n))
cout<else
cout<return 0;
}
int prime(int n)
{int flag=1,i;
for (i=2;iif (n%i==0)
flag=0;
return(flag);
}

4.4 题
#include
using namespace std;
int main()
{int fac(int);
int a,b,c,sum=0;
cout<<"enter a,b,c:";
cin>>a>>b>>c;
sum=sum+fac(a)+fac(b)+fac(c);
cout<return 0;
}
int fac(int n)
{int f=1;
for (int i=1;i<=n;i++)
f=f*i;
return f;
}

4.5 题
#include
#include
using namespace std;
int main()
{double e(double);
double x,sinh;
cout<<"enter x:";
cin>>x;
sinh=(e(x)+e(-x))/2;
cout<<"sinh("<return 0;
}
double e(double x)
{return exp(x);}

4.6 题
#include
#include
using namespace std;
int main()
{double solut(double ,double ,double ,double );
double a,b,c,d;
cout<<"input a,b,c,d:";
cin>>a>>b>>c>>d;
cout<<"x="<return 0;
}
double solut(double a,double b,double c,double d)
{double x=1,x0,f,f1;
do
{x0=x;
f=((a*x0+b)*x0+c)*x0+d;
f1=(3*a*x0+2*b)*x0+c;
x=x0-f/f1;
}
while(fabs(x-x0)>=1e-5);
return(x);
}

4.7 题
#include
#include
using namespace std;
int main()
{void godbaha(int);
int n;
cout<<"input n:";
cin>>n;
godbaha(n);
return 0;
}
void godbaha(int n)
{int prime(int);
int a,b;
for(a=3;a<=n/