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I am not able to understand what we have to do in this question. Can anyone help me to understand the problem and how to approach it with python?
You have to find x which will give correct equation. For simple example you have equation x-5=0, so you have to find x which will give 0, i.e. 5-5=0. About approach it looks like we have range 0-100, medium is 50, 50-5=45 it's more than 0. Next now we have range 0-50, and so on. Sorry for my bad english)
But how do I implement this thing in a code. This way I can get a single solution only
"here A, B and C all would be positive so that function is monotonic. Solution is guaranteed to exist somewhere in range 0 <= x <= 100."
You have begin and end point begin=0, end=100. In the loop while begin < end you took middle i.e. begin+end/2, take result with whis x=middle,
if result more than 0 then end=middle-1, else begin=middle+1, and do it while you result!=0;
if you are asking about how to do loop:
in C++
int input;
cin>>input;
for(int i=0;i< input;i++){
int A,B,C;
cin>>A>>B>>C;
And here do binary search x 0-100 with A B C
}
Yes. But used function is monotonic (always increasing), so it have exactly one root.
import math n=int(input()) l=[] for _ in range(n): a,b,c,d=map(float,input().split()) mi=0 ma=100 while(mi<=ma): mid=(mi+ma)/2 x=mid f=ax + b math.sqrt(x ** 3) - c * math.exp( -x / 50) - d if (f>=0): mi=mid+1 else: ma=mid-1 l.append(mid) print(l)
It is not good to post solutions unless you made a private topic.
And it seems that your solution stops too early - this task require more precision. So I am not satisfied with your solution, anyway.
In this question we basically needed to do a binary search for finding the solution. I am doing it. The program also runs till completion but answer are not correct. The values which I need to put in the equation should be between 0 and 100 so are taken. Then why it stops early. And sorry for posting it here I don't know that we have a private option also.
Look, coding like this:
while(mi<=ma):
...
if (f>=0):
mi=mid+1
else:
ma=mid-1
...you in worst case get precision not better than 0.5 = 5e-1.
You need precision 0.0000001 = 1e-7 or better for this task.
Change code which sets new bounds. And maybe loop condition, too.