10341 - Solve It

[Chung Ha, Yun]

Test set is perfect!!...^^;;

But, I recieve WA...again, again...

Please Help Me..

Where is wrong?

My Algorithm is The Bisection.

[c]
#include<stdio.h>
#include<math.h>

#define Napier 2.71828

int main()
{
int p, q, r, s, t, u;
double start, middle, end;
double start_result, middle_result, end_result;
while(scanf("%d %d %d %d %d %d", &p, &q, &r, &s, &t, &u) > 0)
{
start = 0, middle = 0.5, end = 1;
if(p == 0 && q == 0 && r == 0 && s == 0 && t == 0 && u == 0)
printf("No solution\n");
else
{
while(1)
{
start_result = p*pow(Napier, -1*start) + q*sin(start) + r*cos(start) + s*tan(start) + t*pow(start, 2) + u;
middle_result = p*pow(Napier, -1*middle) + q*sin(middle) + r*cos(middle) + s*tan(middle) + t*pow(middle, 2) + u;
end_result = p*pow(Napier, -1*end) + q*sin(end) + r*cos(end) + s*tan(end) + t*pow(end, 2) + u;

if((middle_result < 0 ? middle_result*-1 : middle_result) <= 0.000000001)
{
printf("%.4lf\n", middle);
break;
}
else if((start_result < 0 && middle_result > 0) || (start_result > 0 && middle_result < 0))
{
end = middle;
middle = (start + end) / 2;
}
else if((end_result < 0 && middle_result > 0) || (end_result > 0 && middle_result < 0))
{
start = middle;
middle = (start + end) / 2;
}
else
{
printf("No solution\n");
break;
}
}
}
}
return 0;
}
[/c]

[shahriar_manzoor]

Can't you write e=exp(1);

[Chung Ha, Yun]

I modified something..

So, I recieve Accepted..

Thanks~*

[Dominik Michniewski]

Could anyone tell me, what could be wrong with this code ? I got WA ....

Code: Select all

cutted off spoiler ...

[Adrian Kuegel]

Ok, your value is not precise enough, finally I found it. Try with step*=0.9 and min = f(1), max = f(0) because the function is monotonic decreasing.

[Leo_Jeru]

{ @BEGIN_OF_SOURCE_CODE }
{ @JUDGE_ID: 18459CN 10341 PASCAL "EA" }
program ProbF(input, output);
const
ep = 1e-14;

var
p, q, r, s, t, u: integer;

function f(x: extended): extended;
begin
f := p * exp(-x) + q * sin(x) + r * cos(x) + s * sin(x) / cos(x) + t * sqr(x) + u;
end;

function f_(x: extended): extended;
begin
f_ := - p * exp(-x) + q * cos(x) - r * sin(x) + s / sqr(cos(x)) + 2 * t * x;
end;

procedure solve;
var
x0, x1: extended;
i: integer;
begin
if(p = 0)and(q = 0)and(r = 0)and
(s = 0)and(t = 0)and(u <> 0)then
begin
writeln(output, 'No solution');
exit;
end;
x1 := 0;
while(abs(f(x1)) < ep)or(abs(f_(x1)) < ep)do x1 := x1 + 1;
for i := 1 to 5000 do
begin
x0 := x1;
{if abs(f_(x0)) < ep then
begin
writeln(output, 'No solution');
exit;
end;}
x1 := x0 - f(x0) / f_(x0);
if abs(x1 - x0) < ep then break;
end;
if abs(f(x1)) < ep then
writeln(output, x1:0:4)
else
writeln(output, 'No solution');
end;

begin
while not eof(input) do
begin
readln(input, p, q, r, s, t, u);
solve;
end;
end.
{ @END_OF_SOURCE_CODE }

[Helal Md. Morshed Alam]

TRY THE CONCEPT OF BINARY SEARCH.

[amd-RS]

Why I get WA ??? For the test case it works ...

My code here:

[c]/* @JUDGE_ID: 13213RX 10341 C "<a href="http://cpu.ucpel.tche.br/">Clube de Programa

[Observer]

Excuse me.

Is the given function strictly increasing/decreasing?

Note: "the given function" refers to
f(x) = p*e^(-x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x^2 + u

[hager]

Is the given function strictly increasing/decreasing?

I have never plotted it, but considering all the trigonometric functions I would say no, it is definitely not strictly increasing / decreasing. However, you can be sure that the function has at most one root in between 0 and 1 inclusive.

Best regards

[Observer]

Well...

I think that Bisection Method can be used iff the function is strictly increasing/decreasing within the limits...
Or am I wrong??

[Observer]

Well...

I think that Bisection Method can be used iff the function is strictly increasing/decreasing within the limits...
Or am I wrong??

OK. I see that I'm wrong. It should be:
Bisection Method can be used iff f(upper) and f(lower) are of opposite signs...

Anyway, I've just got an ACC!!!!!!! Thx everyone!!!

[Alexander Kozlov]

Please, help! What wrong with this code?

Code: Select all

#include <iostream.h>
#include <math.h>
#include <stdio.h>

int p,q,r,s,t,u;
double a, b, c;

double f(double x)
{
  return p/exp(x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x*x + u;
}

int main() {

  while ((cin >> p >> q >> r >> s >> t >> u) != NULL) {

    a = 0.0; b = 1.0;

    if (f(b) > 0 || f(a) < 0) {
      printf("No solution\n");
      continue;
    }

    while (b - a >= 0.00001) {

      if (f(a) == 0) {
        printf("%.4lf\n", a);
        break;
      }

      if (f(b) == 0) {
        printf("%.4lf\n", b);
        break;
      }

      c = (a + b) / 2;

      if (f(a) * f(c) > 0) a = c; else b = c;

    }

    printf("%.4lf\n", c);

  }

  return 0;
}

[Noim]

i am not sure about the bugs of your problem but you can check your prgram for the inputs when the answer will be 0.0000 or 1.0000.

[anupam]

actually nr or bisection method need not know the fn is either inc or dec.
if you check two limits for which the funtion gives values of opposite sign, you can be sure that there is a sol between the limits.