用分而治之的方法解决问题 [英] solve the problem with divide and conquer approach

问题描述

hi
如何计算小数点后最多4位的x数量？ p，q，r，s，t，u是问题的输入

（0< = x< = 1，我们知道这个函数在0< = x< = 1下降了）

hi how i can calculate the amount of x up to 4 digits after decimal? p , q , r , s ,t , u are the inputs of problem
( 0<=x<=1 and we know this function in 0<=x<=1 is descending)

p×e^(-x)+q×sin(x)+r×cos(x)+s×tan(x)+t×x^2+u=0

请帮帮我。

非常感谢。

please help me.
thanks alot.

推荐答案

这是一个数学问题：这个函数不是线性的，还有其他的困难。所以D& C不是一个解决方案，但问题的一部分：-O



如果你正在寻找一个4位数的解决方案，我会像这样运行一个大循环：

元代码：

At first this is a mathematical issue: this function isnt linear and has other dificulties. So D&C isnt a solution but part of the problem :-O

If your are looking for a 4-digit solution, I would ran a big loop like this:
Meta code:

float results[10000];
int n = 0;
for( float step = 0; i < 1; i+=0.00001 )
{
   results[n] = callProblem( i );
}
//take a look on the sample

一些分而治之的方法

some divide and conquer approach

float x = 0;
float diff = 1;//Startvalue
while( )
{
   result = callProblem(x);

  if( result > 0 )
{
x -= diff;
diff = diff/2;
result = callProblem(x);
}
else
{
x += diff;
diff = diff/2;
result = callProblem(x);
}
}

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