c语言二分法求方程的近似解

#include#include#includedouble f(double x){ return x*x*x-3*x+1;}int main(){ double i=1.0, j=2.0; while(fabs(i-j)>=0.001) { if(f((i+j)/2.0)<0) i=(i+j)/2.0; else j=(i+j)/2.0; }; printf("%lf",i); return 0;}