POJ 1905 Expanding Rods (二分+计算几何+精度处理)

标签：acm   c语言   算法   编程   二分   

题目地址：POJ 1905

用二分枚举h，然后判断弧长是否符合条件。重点还是在精度问题上，具体看代码吧。。

#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
using namespace std;
#define LL __int64
#define pi acos(-1.0)
const int mod=9901;
const int INF=0x3f3f3f3f;
const double eqs=1e-6;
bool Judge(double f1, double f2)
{
        return f2-f1>1e-4;
}
bool judge(double f1, double f2)
{
        return fabs(f1-f2)<eqs;
}
double bin_search(double lenth, double L)
{
        double low=0, high=L/2, mid, r;
        while(Judge(low,high)){
                mid=(low+high)/2;
                r=(4*mid*mid+L*L)/(8*mid);
                if(2*r*asin(L/(2*r)) < lenth)
                        low=mid;
                else high=mid;
        }
        return mid;
}
int main()
{
        double L, c, n, lenth;
        while(scanf("%lf%lf%lf",&L,&c,&n)!=EOF&&L>=0){
                lenth=(1+n*c)*L;
                printf("%.3f\n",bin_search(lenth, L));
        }
        return 0;
}

POJ 1905 Expanding Rods (二分+计算几何+精度处理)

标签：acm   c语言   算法   编程   二分   

原文地址：http://blog.csdn.net/scf0920/article/details/43063531