[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。