#include <bits/stdc++.h>
using namespace std;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

typedef __float128 db;

db ta;
db solve(db x){
	//from x to 1
	db M = (ta + 1) * (1 - x) / 2;
	
	//cout << x << " " << 1 << endl;
	//cout << M << endl;
	
	db h = x * ta;	
	db res = h * h / 2;
	
	//from 0 to M y + h
	res += M * M / 2 + h * M;
	//from 0 to 1 - x - M y + 1
	res += (1 - x - M) * (1 - x - M)/2 + (1 - x - M) * ta;
	
	return res;
}



const int N = 1 << 17;
db res[N];
vector<db> ans[N]; 



int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int w, h, n;
	cin >> w >> h >> n;
	ta = 1.0 * h / w;
	
	
	//cout << solve(3./8.) << endl;
	//return 0;
	res[0] = solve(0);
	
	//cout << 8 * 8 *(solve(3./8.) + res[0] * 3/8 * 3/8 + 1./8) << endl;
	//cout << res[0] << endl;
	//return 0;
	FOR(i, 1, n + 1){
		db l = 0, r = 1;
		FOR(t, 0, 200){
			db m1 = (2 * l + r) / 3;
			db m2 = (l + 2 * r) / 3;
			
			db res1 = res[i - 1] * (m1 * m1) + solve(m1);
			db res2 = res[i - 1] * (m2 * m2) + solve(m2);
			
			if(res1 > res2)
				l = m1;
			else
				r = m2;
		}
		res[i] = res[i - 1] * (l * l) + solve(l);
		for(auto x : ans[i - 1])
			ans[i].PB(x * l);
		ans[i].PB(l);
	}
	res[n] += ta * ta / 2;
	//cout << res[n] << endl;
	cout << fixed << setprecision(10);
	
	cout << (long double)(1.0 * res[n] * w * w) << "\n";
	FOR(i, 0, min(n, 10))
		cout << (long double)(ans[n][i] * w) << "\n";
	
	return 0;
}