用d＆c方法求解一个方程

Question

我想用这个方法计算这个方程中x的值为小数点后四位数，并输入p，q，r，s，t，u的值。如何做到这一点？

时间限制：1秒

内存限制：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

我已经广义二分法成未测试递归多段方法：

#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(); 
}