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3 0.222018 23.901887 121.909183 39.096669 110.210922 20.270030 138.355025 2028.716904 25.079551
1.561582 -1 -1
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double PI = 3.1415926535897932384626433832795;
const double eps = 1e-10;
int T;
double x,y,v,ans;
void Search() {
double mid,l,r,vx,vy,t,mVal;
l = 0;r = PI/2.0;
bool suc = false; //判断能不能到达目标点
//二分查找能够到达该点的角度
while(r-l>=eps)
{
mid = (l+r)/2.0;
// 求出初始水平速度和初始竖直速度,以及当横坐标到达x时的时间
vx = v*cos(mid),vy = v*sin(mid);t = x/vx;
//求出横坐标为x时所能到达的Y值,Y>y则说明角度太大,Y<y说明角度太小
mVal = vy*t-0.5*9.8*t*t;
if(mVal < y)
{
l = mid;
}
else if(mVal > y || vy/9.8 < t)// vy/9.8 < t是用来判断 当前的值是不是最小的【PS:由于OJ数据问题,即便不加也不会WA】
{
suc = true;
ans = mid;
r = mid;
}
else
{
suc = true;
ans = mid;
break;
}
}
if(suc) printf("%.6lf\n",ans);
else printf("-1\n");
}
int main () {
//freopen("input.in","r",stdin);
for(scanf("%d",&T);T--;)
{
scanf("%lf %lf %lf",&x,&y,&v);
Search();
}
return 0;
}版权声明:本文为博主原创文章,未经博主允许不得转载。
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原文地址:http://blog.csdn.net/acmore_xiong/article/details/47037557