[二分查找变形]弯曲的木杆(POJ 1905)

这题很有意思，除了一般的程序知识，还用了数学的东西。推导了一些公式什么的。
这次用了两种方法做，层层深入，希望能让大家体会到代码的改良过程
第一步：

#include <iostream>#include <cstdio>#include <cmath>using namespace std;double binary_searsh(double left, double right, double L,double L1){    double mid,L2 = L * L;    while (right - left > 1e-6){        mid = (left + right) / 2; // regard mid as h         double rt = (4 * mid * mid + L2)/ (8 * mid);        double rR = 2*rt *asin(L / (2 * rt));        if (rR < L1){            left = mid;        } else {            right = mid;        }    }     return mid;}int main(){    double h = 0,L,n,c;    while (cin >> L>> n>> c&& c != -1){        h = 0;        if (L == 0 || n == 0 || c == 0){            printf("%.3lf\n",0);        } else {            double L1 = (1 + n * c) * L;            double ans = binary_searsh(0,L/2, L, L1);            printf("%.3lf\n",ans);        }    }}

改良：

#include <iostream>#include <cstdio>#include <cmath>using namespace std;double binary_searsh(double left, double right, double L,double L1){    double mid,L2 = L * L;    while (right - left > 1e-6){        mid = (left + right) / 2; // regard mid as h         double r_2 = mid + L2 / (4 * mid);        double rR = r_2 * asin(L / (r_2));        if (rR < L1){            left = mid;        } else {            right = mid;        }    }     return mid;}int main(){    double h = 0,L,n,c;    while (cin >> L>> n>> c&& c != -1){        h = 0;        if (L == 0 || n == 0 || c == 0){            printf("%.3lf\n",0);        } else {            double L1 = (1 + n * c) * L;            double ans = binary_searsh(0,L/2, L, L1);            printf("%.3lf\n",ans);        }    }}

改进部分在二分搜索部分，表达式进行处理之后，进行运算次数减少。

总结：
输出精度问题，还是用printf解决的好。