二元搜索解优化问题

#include <stdio.h> //binSearch计算x^2=2的近似解，精度1e-5; 证明二元搜索的快速低复杂度寻优能力 float func(float x) { return x * x - 2; } float binSearch(float beg, float end, float acc) { float mid; while ( end - beg > acc) { mid = (end + beg) / 2; if (func(mid) > 0) { //递增是大于，否者相反 end = mid; } else { beg = mid; } printf("beg,end,acc %f,%f,%f\n",beg,end,acc); } return mid; } int main () { float a = binSearch(1, 2, 1e-5); printf("x*x = 2 solution is %f\n", a); //binSearch(1, 2, 1e-2) printf("x*x = 2 solution is %f\n", binSearch(1, 2, 1e-6)); printf("x*x = 2 solution is %f\n", binSearch(1, 2, 1e-10)); //1e-10明显超出float精度了，造成算法不收敛； return 0; } 圆剖面，倒水之后水面下和总半圆面积比值r（0.6比如），问灌水高度多少；转为优化问题 灌水高度h介于[0,R]之间优化；