[二分] [计算几何] AtCoder Beginner Contest 144 D Water Bottle

题意：给定一底面积为正方形的方体并装入x体积的水，问最大倾斜角度使得水不流出

如图所示只有两种情况且面积具有单调性 二分求解即可

/*
    Zeolim - An AC a day keeps the bug away
*/
  
//#pragma GCC optimize(2)
//#pragma GCC ("-W1,--stack=128000000")
#include <bits/stdc++.h>
using namespace std;
#define mp(x, y) make_pair(x, y)
#define fr(x, y, z) for(int x = y; x < z; ++x)
#define pb(x) push_back(x)
#define mem(x, y) memset(x, y, sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair <int, int> pii;
typedef std::vector <int> vi;
void re(ll &x){x=0;char s=getchar();while(s<'0'||s>'9')s=getchar();while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}}
void wr(ll x){if(x>9) wr(x/10);putchar(x%10+'0');}
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 386910137;
const ull P = 13331;
const int MAXN = 1e6 + 100;
 
ld a, b, x, y;
 
ld to(ld x) { return (x / 180.0) * PI; }
ld ret(ld x) { return x / PI * 180.0; }
 
ld cal(ld deg)
{
	if(deg >= y)
	{
		ld rb = a * tan(PI / 2.0 - deg) ;
		return a * b * a - a * rb / 2.0 * a;
	}
	else
	{
		ld ra = b * tan(deg);
		return (b * ra / 2.0 * a);
	}
}
 
 
int main()
{  
    //ios::sync_with_stdio(0);
    //cin.tie(0); cout.tie(0);
    //freopen("d:out.txt","w",stdout);
    //freopen("d:in.txt","r",stdin);  
  	
  	cin >> a >> b >> x;
  	
  	y = atan(a/b);
  	
  	ld fst = 0, lst = PI / 2.0, mid;
  	
  	for(int i = 0; i < 1000; ++i)
  	{
  		mid = (fst + lst) / 2.0;
  		
  		if(cal(mid) >= x)
  			lst = mid;
  		else
  			fst = mid;
	}
	  
	cout << fixed << setprecision(10) << 90.0 - mid / PI * 180.0 << '
';
    
    return 0;
}