Solving an equation with d&c method - c++

Question

I want to calculate value of x in this equation to 4 digits after the decimal with divide and conquer method.Values of p,q,r,s,t,u are input. How to do it in?
Time limits: 1 sec
Memory limits: 64 MB
float results[10000];
int n = 0;
for( float step = 0; i < 1; i+=0.00001 )
{
results[n] = callProblem( i );
}
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);
}
}

Answer 1

I have generalized the bisection method into an untested recursive multi-section method:
#include <math.h>
#include <stdio.h>
#include <time.h>
#define P 1.0
#define q 2.0
#define r 3.0
#define s 1.0
#define t 5.0
#define u -6.0
// number of sub-intervals in search interval
#define INTERVALS 10
// accuracy limit for recursion
#define EPSILON 1.0e-4
double f(double x) {
double y = P * exp(-x) + q*sin(x) + r*cos(x) + s*tan(x) + t*x*x + u;
return y;
}
// sign macro: -1 for val < 0, +1 for val > 0
#define SGN(val) ((0.0 < val) - (val < 0.0))
// return approximate x for function(x)==0.0
// "function" points to the function to be solved
double solve(double xMin, double xMax, double (*function)(double)) {
double arguments[INTERVALS + 1];
double values[INTERVALS + 1];
int prevSign;
int sign;
if (fabs(xMax - xMin) < EPSILON) {
// interval quite tight already
return (xMax + xMin) / 2.0;
}
// split the interval into sub-intervals
// evaluate the function for INTERVALS+1 equidistant points
// across our search interval
for (int i = 0; i <= INTERVALS; i++) {
double x = xMin + i*((xMax - xMin) / INTERVALS);
arguments[i] = x;
values[i] = function(x);
}
// look for adjacent intervals with opposite function value signs
// if f(Xi) and f(Xi+1) have different signs, the root must be
// between Xi and Xi+1
prevSign = SGN(values[0]);
for (int i = 1; i <= INTERVALS; i++) {
sign = SGN(values[i]);
if (sign * prevSign == -1) {
// different signs! There must be a solution
// Shrink search interval to the detected sub-interval
double x = solve(arguments[i - 1], arguments[i], function);
return x;
}
// remember sign for next round
prevSign = sign;
}
// no solution found: return not-a-number
return NAN;
}
int main(unsigned argc, char **argv) {
clock_t started = clock();
clock_t stopped;
double x = solve(0.0, 1.0, &f);
if (isnan(x)) {
printf("\nSorry! No solution found.\n");
} else {
printf("\nOK! Solution found at f(%f)=%f\n", x, f(x));
}
stopped = clock();
printf("\nElapsed: %gs", (stopped - started) / (double)CLOCKS_PER_SEC);
// wait for user input
getchar();
}