BestCoder22 1003.NPY and shot 解题报告

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5144

题目意思：有个人抛物体，已知抛的速度和高度，问可以抛到的最远距离是多少。即水平距离。

　　做的时候是抄公式的，居然过了，幸运幸运............

　　

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;

const double g = 9.8;

int main()
{
    int T;
    double h, v;
    while (scanf("%d", &T) != EOF)
    {
        while (T--)
        {
            scanf("%lf%lf", &h, &v);
            double tmp = sqrt(v*v + 1ll * 2 * g * h);
            double t = tmp / g;
            double ans = t * v;
            printf("%.2lf
", ans);
        }
    }
    return 0;
}

　　乌冬子做的比较正规，是用三分做的，还有浮点精度控制，值得学习 ^_^  （未经他允许就盗他版权，应该不会怪我的......）

　　

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <map>
#include <numeric>
#include <vector>

using namespace std;

const double EPS = 1e-8;
const double PI = acos(-1.0);
const double G = 9.8;

int h, v;

double F(double ang)
{
  double hv = cos(ang)*v;
  double vv = sin(ang)*v;
  double t1 = vv/G;
  double t2 = sqrt(2*(vv*t1-G*t1*t1/2.0+h)/G);
  return hv*(t1+t2);
}

int main()
{
  int kase;
  scanf("%d", &kase);
  while (kase--) {
    scanf("%d%d", &h, &v);
    double l = 0.0, r = PI/2;
    while (l < r-EPS) {
      double d = (r-l)/3.0;
      double ll = l+d;
      double rr = r-d;
      if (F(ll) < F(rr))
        l = ll;
      else
        r = rr;
    }
    printf("%.2f
", F(l));
  }
  return 0;
}