02:二分法求函数的零点

 1 #include<stdio.h>
 2 #include<math.h>
 3 double f(double x)
 4 {
 5     double ans=0;
 6     ans=pow(x,5)-15*pow(x,4)+85*pow(x,3)-225*pow(x,2)+274*x-121;
 7     return ans;
 8 }
 9 int main(int argc, char *argv[])
10 {
11     double x,y,mid,fx,fy,fm;
12     x=1.5;
13     y=2.4;
14     mid=(x+y)/2;
15     fx=f(x);
16     fy=f(y);
17     fm=f(mid);
18     while( fabs(fm-0) > 1e-6)
19     {
20         if(fx*fm<0) { y=mid; fy=f(y); }
21         else { x=mid; fx=f(x); }
22         mid=(x+y)/2;
23         fm=f(mid);
24     }
25     printf("%.6lf\n",mid);
26     return 0;
27 }