#include<stdio.h>
#include<math.h>
#include <stdlib.h>
int main()
{
    float f(float x);
    float a,b,c,jqd;
    float abs(float x);
    printf("请输入区间，精确度\n");
    scanf("[%f,%f],%f",&a,&b,&jqd);
    if(f(a)*f(b)>0) printf("该区间无零点\n");
    else
    {
         if(f(a)*f(b)==0)
         {
                         if(f(a)==0)printf("零点为%f\n",a);
                         else printf("零点为%f\n",b);
                         }
         else
                         {
                              if(abs(a-b)<=jqd)printf("零点是%f\n",a);
                              else
                              {
                                  while(abs(a-b)>jqd)
                                  {
                                                   c=(a+b)*2;
                                                   if(f(c)==0) a=c;
                                                   else
                                                   {
                                                        if(f(a)*f(c)<0) b=c;
                                                        else a=c;
                         }
                              }
                                  }
                                                    }

    }
                                                        printf("零点是%f",a);                                      
                                                        system("pause");
                                                        return 0;
}
                                                        
float f(float x)
{
float y;
y=x*x;
return(y);
}
float abs(float x)
{
return (x>0 ? x : -1*x);
}

      

人来啊

f(x)=x*x;
显然f(-1)*(1)也是大于0的,但[-1,1]区间有零点.