#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define beita 0.618
#define e 0.00001
int main()
{
    float x,a=0,b=10,t1,t2,f0,f1,f2;
     t2=a+beita*(b-a);
    f2=t2*t2-10*t2+36;
    t1=a+b-t2;
    f1=t1*t1-10*t1+36;
    while (abs(t1-t2)>e)
    {
    if (f1<=f2)
    b=t2;
    t2=t1;
    f2=f1;
    t1=a+b-t2;
    f1=t1*t1-10*t1+36;
    else
    a=t1;
    t1=t2;
    f1=f2；
    t2=a+beita*(b-a；
     f2=t2*t2-10*t2+36;
     }
     t0=0.5*(t1+t2);
     f=t0*t0-10*t0+36;
        printf("The answer is %f\n",f);
   
    system("pause");
}

彩笔习惯了matlab代码的风格，忽转C代码，虽问题简单却出了很多问题，还望大婶们不要见笑

改好了

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define beita 0.618
#define e 0.00001
int main()
{
    float x,a=0,b=10,t0,t1,t2,f0,f1,f2,k;
     t2=a+beita*(b-a);
    f2=t2*t2-10*t2+36;
    t1=a+b-t2;
    f1=t1*t1-10*t1+36;
    k=fabs(t1-t2);
    while (k>e)
    {
    if (f1<=f2)
    {b=t2;
    t2=t1;
    f2=f1;
    t1=a+b-t2;
    f1=t1*t1-10*t1+36;
    }
    else
    {
    a=t1;
    t1=t2;
    f1=f2;
    t2=a+beita*(b-a);
     f2=t2*t2-10*t2+36;
    }
     }
     t0=0.5*(t1+t2);
     f0=t0*t0-10*t0+36;
        printf("The answer is %f\n",f0);
   
    system("pause");
}