重庆分公司,新征程启航

为企业提供网站建设、域名注册、服务器等服务

C语言用函数实现高斯加法 高斯求积公式c语言

高斯变换,在C语言中实现的思路,知道的请告诉我,详细点越好

#include stdio.h

为阳原等地区用户提供了全套网页设计制作服务,及阳原网站建设行业解决方案。主营业务为成都做网站、网站建设、阳原网站设计,以传统方式定制建设网站,并提供域名空间备案等一条龙服务,秉承以专业、用心的态度为用户提供真诚的服务。我们深信只要达到每一位用户的要求,就会得到认可,从而选择与我们长期合作。这样,我们也可以走得更远!

#include iostream.h

#include math.h

#includeiomanip.h

class Angle

{

public:

double degree,cent,second,Hudu,seconds;

//

构造函数

Angle(double _degree,double _cent,double _second)

{

degree=_degree;

cent=_cent;

second=_second;

seconds=(degree*3600+cent*60+second);

Hudu=(degree*3600+cent*60+second)*2*3.1415926535/(360*60*60);

}

Angle()

{

}

SToD()

{

second=seconds-int(seconds/60)*60;

degree=int((seconds-second)/3600);

cent=int((seconds-second-degree*3600)/60);

}

~Angle()

{

}

};

void main()

{

//

正算过程

//

定义变量

Angle B(51,38,43.9023),L(126,2,13.1360),_B,_L;

double L0,l,N,a0,a4,a6,a3,a5,x,y,rou;

rou=(360*60*60)/(2*3.1415926535);

if(int(L.degree)%6!=0)

{

L0=6*(int(L.degree)/6+1)-3+6;

}

else

L0=6*(int(L.degree)/6)-3+6;

l=L.Hudu-L0*3600*2*3.1415926535/(360*60*60);

N=6399698.902-(21562.267-(108.973-0.612*cos(B.Hudu)*cos(B.Hudu))*cos(B.Hudu)*cos(

B.Hudu))*cos(B.Hudu)*cos(B.Hudu);

a0=32140.404-(135.3302-(0.7092-0.0040*cos(B.Hudu)*cos(B.Hudu))*cos(B.Hudu)*cos(B.

Hudu))*cos(B.Hudu)*cos(B.Hudu);

a4=(0.25+0.00252*cos(B.Hudu)*cos(B.Hudu))*cos(B.Hudu)*cos(B.Hudu)-0.04166;

a6=(0.166*cos(B.Hudu)*cos(B.Hudu)-0.084)*cos(B.Hudu)*cos(B.Hudu);

a3=(0.3333333+0.001123*cos(B.Hudu)*cos(B.Hudu))*cos(B.Hudu)*cos(B.Hudu)-0.166666

7;

a5=0.0083-(0.1667-(0.1968+0.0040*cos(B.Hudu)*cos(B.Hudu))*cos(B.Hudu)*cos(B.Hudu)

)*cos(B.Hudu)*cos(B.Hudu);

x=6367558.4969*B.Hudu-(a0-(0.5+(a4+a6*l*l)*l*l)*l*l*N)*sin(B.Hudu)*cos(B.Hudu);

y=(1+(a3+a5*l*l)*l*l)*l*N*cos(B.Hudu);

//结果输出

cout"x="xendl;

cout"y="yendl;

// 反算过程

double b,Bf,Nf,Z,b2,b3,b4,b5,l1;

b=(x/6367558.4969);

Bf=b+(50221746+(293622+(2350+22*cos(b)*cos(b))*cos(b)*cos(b))*cos(b)*cos(b))*0.000

0000001*sin(b)*cos(b);

Nf=6399698.902-(21562.267-(108.973-0.612*cos(Bf)*cos(Bf))*cos(Bf)*cos(Bf))*cos(Bf)*c

os(Bf);

Z=y/(Nf*cos(Bf));

b2=(0.5+0.003369*cos(Bf)*cos(Bf))*sin(Bf)*cos(Bf);

b3=0.333333-(0.166667-0.001123*cos(Bf)*cos(Bf))*cos(Bf)*cos(Bf);

b4=0.25+(0.16161+0.00562*cos(Bf)*cos(Bf))*cos(Bf)*cos(Bf);

b5=0.2-(0.1667-0.0088*cos(Bf)*cos(Bf))*cos(Bf)*cos(Bf);

_B.seconds=Bf*rou-(1-(b4-0.12*Z*Z)*Z*Z)*Z*Z*b2*rou;

l1=(1-(b3-b5*Z*Z)*Z*Z)*Z*rou;

_L.seconds=L0*3600+l1;

_B.SToD();

_L.SToD();

请问如何用C语言编程编写一个实现加法的函数,例如y=3x+1这个函数。

#include stdio.h

double f(double x)

{return(x*3+1);}

void main(){

double x;

scanf("%lf",x);

printf("%lf",f(x));

}

C语言,我想通过调用口令函数和加法函数实现加法运算

#includestdio.h

#includeconio.h

void login();

void add();

bool flags = true;

void main()

{

while(flags)

login();

//add();

//getch();

}

void login()

{

int k;

int L=888;

printf("口令888;请输入验证口令:");

scanf("%d",k);

if(L==k)

{

int choice;

printf("您输入正确,欢迎使用C计算器\n");

printf("+=======选择要做的运算=======+\n");

printf("+= 1、加法 =+\n");

printf("+= 2、减法 =+\n");

printf("+= 3、乘法 =+\n");

printf("+= 4、除法 =+\n");

printf("+= 5、求余 =+\n");

printf("+= 0、退出 =+\n ");

printf("+============================+\n");

scanf("%d",choice);

switch(choice)

{

case 1:

add();

break;

case 0:

flags = false;

break;

default : printf("您输入有误");

break;

}

}

else

{

printf("您输入错误,请重新输入!");

}

}

void add()

{

int jia,fa;

printf("请输入2个数");

scanf("%d %d",jia,fa);

printf("%d+%d=%d\n",jia,fa,jia+fa);

}

首先你add写在login里面,main先执行login在执行add,可是add在login了,所以main保留一个login,函数不能嵌套,add写在外面。为了实现循环,添加一个标志位flags初始为true,在选项里添加一个退出,选择之后flags设为flase,退出计算器。


网站题目:C语言用函数实现高斯加法 高斯求积公式c语言
本文路径:http://cqcxhl.cn/article/ddocsec.html

其他资讯

在线咨询
服务热线
服务热线:028-86922220
TOP