在椭圆问题上制作程序:第2部分重新发布忽略 [英] Making a program on ellipse problem:part 2 Repost Ignore

查看:109
本文介绍了在椭圆问题上制作程序:第2部分重新发布忽略的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

您好,..是问在椭圆形物体上编写程序..."问题的那个人...不幸的是,我的老师没有给我们提供要使用的软件...他想让我们发现它靠我们自己...我有我的代码,但是嗯...我该怎么说...我有点喜欢编程,但是我想学习...



#include
主要(0
{
int n1,n2,n3,n4,n5,n6,n7,n8,n9,n10,n11,n12,n13,x,y;
字符a,b,c,d,e,f,g;
a =(a)*(a);
b =(b)*(b);
c =(c)*(c);
d =(x)*(x);
e =(y)*(y);
f = x;
g = y;
printf("a.找到方程图的x和y截距\ n");
printf("b.找到焦点的后代\ n");
printf("c.找到长轴和短轴的长度\ n");
printf("d.绘制方程图\ n");
printf(输入数字:");
scanf(%d",& n1);
printf(输入另一个数字:");
scanf(%d",& n2);
printf(输入要用作分母的数字:");
scanf(%d",& n3);
1 = n1a/n3 + n2b/n3;
printf(%d%c/%d +%d%c/%d = 1 \ n",n1,a,n3,n2,b,n3);
printf(%c/%d +%c/%d = 1 \ n",d,n4,e,n5);
printf(%c/%d +%c/%d \ n",d,n6,e,n7);
1 = a/n6 + b/n7;
printf("a等于%d,b等于%d \ n",n7,n6);
y = 0;
printf(为x \ n解决");
printf(%c =%d \ n",d,n6);
printf(%c = +-%d \ n",f,n8);
x = 0;
printf(为y \ n解决");
printf(%c =%d \ n",e,n7);
printf(%c =%d \ n",g,n9);
printf(解决c \ n");
(c)*(c)=(a)*(a)-(b)*(b);
printf(%c =%c-%c \ n",c,a,b);
printf(%c =%d-%d \ n",c,n7,n6);
printf(%c =%d \ n",c,n10);
printf(%c = +-%d \ n",c,n10);
printf(焦点为F1(%d,%d)和F2(%d,-%d)\ n",n11,n10,n11,n10);
printf(长轴长度2a =%d \ n",n12);
printf(次轴长度2b =%d \ n",n13);

那就是我的代码结束的地方...它仍然缺少用于绘制图形的代码...这是非常基本的吗?可怜的我...如果我有错误,请告诉我.. .我确定我有...


您已经对这个问题有了答案,请不要再重新发布.

Hi..it''s meh the one who asked the question"Making a program on ellipse thing..."..unfortunately my instructor did not gave us what software to use...he wants us to discover it on our own...I have my code but hmm...how should i say this...i kinda suck in programming but i wanna learn...



#include
main(0
{
int n1,n2,n3,n4,n5,n6,n7,n8,n9,n10,n11,n12,n13,x,y;
char a,b,c,d,e,f,g;
a=(a)*(a);
b=(b)*(b);
c=(c)*(c);
d=(x)*(x);
e=(y)*(y);
f=x;
g=y;
printf("a. Find the x and y intercepts of the graph of the equation\n");
printf("b. find the coodinates of the foci\n");
printf("c. Find the length of the major axes and the minor axes\n");
printf("d. Sketch the graph of the equation\n");
printf("Enter a number:");
scanf("%d",&n1);
printf("Enter another number:");
scanf("%d",&n2);
printf("Enter a number to be used as denominator:");
scanf("%d",&n3);
1=n1a/n3+n2b/n3;
printf("%d%c/%d+%d%c/%d=1\n",n1,a,n3,n2,b,n3);
printf("%c/%d+%c/%d=1\n",d,n4,e,n5);
printf("%c/%d+%c/%d\n",d,n6,e,n7);
1=a/n6+b/n7;
printf("a is equal to %d and b is equal to %d\n",n7,n6);
y=0;
printf("Solve for x\n");
printf("%c=%d\n",d,n6);
printf("%c=+-%d\n",f,n8);
x=0;
printf("Solve for y\n");
printf("%c=%d\n",e,n7);
printf("%c=%d\n",g,n9);
printf("Solve for c\n");
(c)*(c)=(a)*(a)-(b)*(b);
printf("%c=%c-%c\n",c,a,b);
printf("%c=%d-%d\n",c,n7,n6);
printf("%c=%d\n",c,n10);
printf("%c=+-%d\n",c,n10);
printf("The foci are F1(%d,%d) and F2(%d,-%d)\n",n11,n10,n11,n10);
printf("Major axis length 2a=%d\n",n12);
printf("Minor axis length2b=%d\n",n13);

That''s where my code ends...it still lacks the code for the sketching of the graph...it''s very basic isn''t it?poor me...tell me if i have errors...im sure i have...


You already have an answer to this question, please don''t just repost the same again.

推荐答案

写道:

如果我有错误,请告诉我

tell me if i have errors



通过编译器运行它会发生什么?



What happens when you run it through the compiler?


这篇关于在椭圆问题上制作程序:第2部分重新发布忽略的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆