Problem:
Score : 400 points

Problem Statement

Takahashi has a water bottle with the shape of a rectangular prism whose base is a square of side a~\mathrm{cm} and whose height is b~\mathrm{cm}. (The thickness of the bottle can be ignored.)

We will pour x~\mathrm{cm}^3 of water into the bottle, and gradually tilt the bottle around one of the sides of the base.

When will the water be spilled? More formally, find the maximum angle in which we can tilt the bottle without spilling any water.

Constraints

All values in input are integers.

1 \leq a \leq 100

1 \leq b \leq 100

1 \leq x \leq a^2b

Input

Input is given from Standard Input in the following format:

a b x

Output

Print the maximum angle in which we can tilt the bottle without spilling any water, in degrees.
Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

2 2 4

Sample Output 1

45.0000000000

This bottle has a cubic shape, and it is half-full. The water gets spilled when we tilt the bottle more than 45 degrees.

Sample Input 2

12 21 10

Sample Output 2

89.7834636934

This bottle is almost empty. When the water gets spilled, the bottle is nearly horizontal.

Sample Input 3

3 1 8

Sample Output 3

4.2363947991

This bottle is almost full. When the water gets spilled, the bottle is still nearly vertical.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdlib>
#include <iostream>
#include <string>
#include <time.h>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <iomanip>
#include <vector>
#include <functional>
#include <numeric>
#include <cassert>
#include <algorithm>
#include <functional>
#include <tuple>
#include <bitset>
#include <queue>
#include <map>
//#include "bits/stdc++.h"
using namespace std;
//using namespace std::vector;
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define M_PI 3.14159265358979323846
#define huge 1000000
typedef long long int ll;
ll gcd(ll a, ll b);
ll lcm(ll x, ll y);
ll bur(ll N, ll X);
std::vector<ll> enum_div(ll n);
long long comb(int n, int r);
bool IsPrime(int num);
bool dp[110][10010];
class UnionFind;
using Graph = vector<vector<int>>;
ll co[300][300] = {};

//エラトステネスの篩
vector<bool> is_prime(huge);
vector<ll> P;
void Eratosthenes(const ll N)
{
	for (ll i = 0; i <= N; i++)
	{
		is_prime[i] = true;//初期化
	}
	for (ll i = 2; i <= N; i++)
	{
		if (is_prime[i])
		{
			for (ll j = 2 * i; j <= N; j += i)
			{
				is_prime[j] = false;
			}
			P.emplace_back(i);
		}
	}
}

//UF木
class UnionFind {
public:
	//親
	vector<int> Parent;

	//初期状態
	UnionFind(int N) {
		Parent = vector<int>(N, -1);
	}

	int root(int A) {
		if (Parent[A] < 0) return A;
		return Parent[A] = root(Parent[A]);
	}

	//サイズ確認
	int size(int A) {
		return -Parent[root(A)];
	}

	int connect(int A, int B) {
		//root同氏をくっつける
		A = root(A);
		B = root(B);

		//結合済み
		if (A == B) {
			return false;
		}
		//大きいほうに小さいほうをくっつける
		if (size(A) < size(B)) swap(A, B);

		//Aのサイズ更新
		Parent[A] += Parent[B];
		//Bの親をAに
		Parent[B] = A;

		return true;
	}

};

int main(void) {

	float N = 0;
	float A = 0;
	float B = 0;
	ll C = 0;
	ll D = 0;
	//ll G = 0;
	ll M = 0;
	ll L = 0;
	ll Q = 0;
	string S, T, U[52];
	char fi[50], fv[50];
	char board[4][4];
	ll count = 0;
	double count2 = 0;
	float n = 0;
	ll min = 300000;
	float max = 0, max2 = 0, max3 = 0, max4 = 0;
	float dist1;
	float dist2;
	float dist3 = 1;
	ll dist4 = 0;
	float now = 0;
	int tmax = 10000;
	int tmax2 = 0;
	//ll count = 0;
	//int flag = 0;
	//char temp, head, state;
	//int is = 0;
	bool start = false;
	ll left = 0, right = 0;
	cin >> A >> B >> N;
	

	float hena = 0,henb=0,henc=0;
	//長方形面積
	dist1 = (float)A*(float)B;
	//水の面積
	dist2 = (float)N / (float)(A);
	dist3 = (float)dist2 / (float)dist1;

	//cout << dist3 << " " << dist2 << endl;

	if (dist3 <= 0.5) {
		hena = (float)B;
		henb = N / A * 2 / (float)B;
		
		//cout << hena << endl;

		//cout << henb << endl;
		float degree = atan(hena / henb) *180.0 / M_PI;
		cout << fixed << setprecision(10) << degree  << endl;

	}
	else {
		hena = (float)A;
		henb = (A*B - N/A) * (float)2 / (float)A;
		
		float degree = atan(2*(A*A*B-N)/(A*A*A)) *180.0 / M_PI;
		//cout << degree << endl;

		cout << fixed << setprecision(10) << degree << endl;
	}
	return 0;
}


std::vector<ll> enum_div(ll n)//nの約数を列挙
{
	vector<ll> res;
	for (ll i = 1; i*i <= n; ++i) {
		if (n % i != 0) continue;
		res.push_back(i);
		if (n / i == i) continue; // 上の行で追加済み。
		res.push_back(n / i);
	}
	return res;
}
ll bur(ll N, ll X) {
	ll pan = 1;
	ll pat = 1;
	ll burg = 0;



	if (X > burg) {
		if (N == 0)
			return pat;


		burg += pan;
		burg += bur(N - 1, X);
		burg += pat;
		burg += bur(N - 1, X);
		burg += pan;


	}

	return burg;

}

ll gcd(ll a, ll b) {

	/* 自然数 a > b を確認・入替 */
	if (a<b) {
		ll tmp = a;
		a = b;
		b = tmp;
	}
	ll x = b;
	ll y = a;
	/* ユークリッドの互除法 */
	ll r = a % b;
	while (r != 0) {
		a = b;
		b = r;
		r = a % b;
	}


	return y / b*x;
}





bool IsPrime(int num)
{
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			// 素数ではない
			return false;
		}
	}

	// 素数である
	return true;
}

// 最大公約数を求める関数
ll lcm(ll x, ll y)
{
	if (x == 0 || y == 0)  // 引数チェック
	{
		//fprintf(stderr, "引数エラーです。\n");
		return 0;
	}

	return (x * y / gcd(x, y));
}

//combination
long long comb(int n, int r) {
	std::vector<std::vector<long long>> v(n + 1, std::vector<long long>(n + 1, 0));
	for (int i = 0; i < v.size(); i++) {
		v[i][0] = 1;
		v[i][i] = 1;
	}
	for (int j = 1; j < v.size(); j++) {
		for (int k = 1; k < j; k++) {
			v[j][k] = (v[j - 1][k - 1] + v[j - 1][k]);
		}
	}
	return v[n][r];
}