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

//#define f(x) (x+sin(x)-1)
#define f(x) ((x)*(x))
//#define f(x) ((x)*exp(x)*(x)*exp(x))

int main() {
	int count = 1000 , i=0;
	double precision = 1e-10 ;
	//num為選擇解哪個方程式 
	double mid , lower=-20 , upper=10, num=2;
	printf("迭代\t區間\t\t\t\t中間值\n");
	printf("-------+-------------------------------+-------------\n");
	
	if(f(num)==(num+sin(num)-1)){
		while (upper - lower > precision && i < count ) {
	    mid = (lower + upper) / 2.0;  

	    printf("#%d\t[%.10f,%.10f]\t解=%.10f\n",i+1,lower,upper,mid);
	    
	    if (fabs(f(mid)) < precision) {
        	break;
        }
        	
	    i++;
	    if ( f(mid) * f(lower) < 0) {
	    	upper = mid;
	    } else {
	    	lower = mid;
	    	}
		}
	}
	
	if(f(num)==(num*num)){
		while (upper - lower > precision && i < count ) {
	    mid = (lower + upper) / 2.0;  
	    printf("#%d\t[%.10f,%.10f]\t解=%.10f\n",i+1,lower,upper,mid);
	    printf("\nf(mid)=%.10f, f(lower)=%.10f, f(upper)=%.10f\n",f(mid),f(lower),f(upper));
	    
	    if (fabs(f(mid)) < precision) {
        	break;
        }
        	
	    i++;
	    if ( f(mid) > precision) {
	    	if(mid>0){
	    		upper = mid;
	    } else if(mid<0){
				lower = mid;
				}
	    	}
		}
	}
	
	if(f(num)==((num)*exp(num)*(num)*exp(num))){
		while (upper - lower > precision && i < count ) {
	    mid = (lower + upper) / 2.0;  
	    printf("#%d\t[%.10f,%.10f]\t解=%.10f\n",i+1,lower,upper,mid);
//	    printf("mid=%.10f,f(mid)=%.10f\n",mid,f(mid));
	    if (fabs(f(mid)) < precision) {
        	break;
        }
        	
	    i++;
	    if ( f(mid) > precision) {
	    	if(mid>0){
	    		upper = mid;
	    } else if(mid<0){
				lower = mid;
				}
	    	}
		}
	}
		
	printf("\n方程式的解為%f\n",mid);
	printf("\n驗算結果f(mid)的值為%f",f(mid));
	
	return 0;
}