100分求救!急急急!!!!c语言求解2元2次方程,求大神啊!

2025年04月03日 05:36
有4个网友回答
网友(1):

这是TI10年的比赛D题,我有详细代码,留邮箱我发给你

网友(2):

for(x=0;x<=450;x++){

for(y=0;y<=450;y++){
if((sqrt(x*x+y*y)-sqrt(x*x+(450-y)*(450-y))==340*num1)&&(sqrt(x*x+y*y)-sqrt((450-x)*(450-x)+y*y)==340*num2))
{
printf("%d %d",x,y);
}
穷举法肯定能算出来,如果整数不行再试小数

网友(3):

sqrt(x*x+y*y)-sqrt(x*x+(450-y)(450-y))=340*num1; 公式1

sqrt(x*x+y*y)-sqrt((450-x)(450-x)+y*y)=340*num2; 公式2

公式2,假设y=0,则

abs(x)-abs(450-x)=340*num2;

1) 0<0x<450: x0=(340*num2+450)/2。且必须要求在(0,450)

2) x>=450: 340*num2-450=0必须要成立。此种情况y始终为0,不考虑

3) x<=0: num2+450必须要成立。此种情况y始终为0,不考虑

公式1,假设x=0,则同理推论得到:

0<0x<450: y0=(340*num1+450)/2。且必须要求在(0,450)


第1轮:x=x0, y=0。x带入公式1,得到new_y。并得到y差。(这里用二分法求一元函数的解)

第2轮:y=new_y。y带入公式2,得到new_x。并得到x差。

重复前2轮。要求y差,x差要收敛。直到y差,x差都很低,结束。

网友(4):

和圆的差不多!晕,,x,y,r都可变!!

先算

sqrt(t1)-sqrt(t2)=sqrt(t12);

 

再代入算!!! 

 

=========================================================

说细点,用参数方程解!!!就简单多了!!

下面只说原理!!!

比如原的参数方程由(x*x+y*y=r*r)变为x=cosA*r,y=sinA*r;

也就是说只需要两个数组就OK了!cos[360],sin[360];

而楼主的问题好比是:

解开

a=(secA)4*r2

b=(tanA)4*r2

a=x2+(1-y)2,

b=(1-x)2+y2,

a-b=2x-2y;

再来一次!!(当然也可以就用上面的方式解开)

x-y=(a-b)\2;

x=(a-b)2\4;y=(a-b)2\4;

........

接下来可以带入算!!!!!!!就a,b,r!!!

 

(最好是精简为三角函数的多少次幂,直接就把它们用数组存起来,大大的减少运算次数!将角度值A循环个360次,当过了360返0就OK了,这时候的x,y的一系列就出来了!)

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