UCV2013I Spoj

Logic :

     For this problem only a image is enough to understand logic. 
     Which is

Source: Click Here

now we can apply sine formula to find solution.

Below you can see code also: 

#include <iostream>

#include <sstream>

#include <cstdio>

#include <cmath>

#include <cstring>

#include <cctype>

#include <string>

#include <vector>

#include <list>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <algorithm>

#include <functional>

using namespace std; 

#define DEBUG(x) cout << '>' << #x << ':' << x << endl;

#define REP(i,n) for(int i=0;i<(n);i++)

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)

#define FORD(i,a,b) for(int i=(a);i>=(b);i--)

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }

const int INF = 1<<29;

typedef long long ll;

inline int two(int n) { return 1 << n; }

inline int test(int n, int b) { return (n>>b)&1; }

inline void set_bit(int & n, int b) { n |= two(b); }

inline void unset_bit(int & n, int b) { n &= ~two(b); }

inline int last_bit(int n) { return n & (-n); }

inline int ones(int n) { int res = 0; while(n && ++res) n-=n&(-n); return res; }

template<class T> void chmax(T & a, const T & b) { a = max(a, b); }

template<class T> void chmin(T & a, const T & b) { a = min(a, b); }

/////////////////////////////////////////////////////////////////////

int main()

{

while(true) {

double r;

double N;

scanf("%lf%lf", &r, &N);

if(r <= 0 || N <= 0) break;

printf("%.2lf\n", r/sin((acos(-1) * 0.5) / N));

}

    return 0;

}

Happy Coding :)