UVa 10341 - Solve It

10341 - Solve It

 * http://acm.uva.es/p/v103/10341.html
 * http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=24&problem=1282

Summary
Given an equation: f(x) = p*e-x + q*sin(x) + r*cos(x) + s*tan(x) + t*x2 + u = 0 and the values for p,q,r,s,t,u.

Find the value for x (if exists), where 0 <= x <= 1.

Explanation
The problem asks for the root of an equation. Since f is a non-increasing function on [0,1], it is sufficient to check the signs for f(0) and f(1) to determine whether there is a root in range [0,1]. There are at least three ways to find a root:


 * The simplest is the Bisection method. It took ~32 ms to answer all the judge's input.


 * A more effective way is the Secant method. It took ~20 ms to answer all the judge's input.


 * The most effective way is the Newton's method. It took ~12 ms to answer all the judge's input and it only requires at most 5 iterations to find the root.

Gotchas

 * Make sure that the root is computed with good precision (EPSILON <= 1e-7 is enough).

Implementations
The easiest to implement is the Bisection method:

A more effective root finding using Secant method:

The most effective is the Newton's method:

Input
0 0 0 0 -2 1 1 0 0 0 -1 2 1 -1 1 -1 -1 1 0 0 0 0 0 0 1 -20 3 -20 -5 6 2 -20 3 -20 -5 6 3 -20 3 -20 -5 6 4 -20 3 -20 -5 6 5 -20 3 -20 -5 6 6 -20 3 -20 -5 6 3 -4 1 -3 -2 5 6 -11 8 -20 -3 1 4 -4 4 -4 -4 5 17 -6 2 -8 -1 3 16 -1 12 -2 -12 4 4 -9 10 -2 -4 8 4 -15 19 0 -5 6 16 -1 9 -11 -14 18 11 -18 11 -7 -6 16 12 -3 1 -4 -18 19 12 -2 10 -3 -8 1 6 -15 11 -19 -7 -13 9 -7 10 0 -4 -1 15 -1 18 -8 -7 5 20 -6 6 -12 -18 -14 5 -5 3 -10 -15 -1 12 -15 18 -9 -4 -19 0 0 0 0 1 -1 0 0 0 0 -1 1

Output
0.7071 No solution 0.7554 0.0000 0.2347 0.2521 0.2689 0.2850 0.3005 0.3154 0.7863 0.3807 0.8016 0.7628 No solution No solution No solution 0.9580 0.8911 0.9489 0.9074 0.0976 0.9066 0.9991 0.2879 0.2879 0.2879 1.0000 1.0000