/*
非常值得重新重做的概率题,化连续为离散,分为(2^M +1)个区间,double型的赌注,则用该区间内的int型来代替,反正效果一样(最终得到的概率是相等的)
等到要输出prv[i]时,再去找在dp数组中,该本金对应的是(2^M+1)个区间中的哪个区间
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <iomanip>
using namespace std;
typedef long long ll;
const int MAX_M = 15;
int M, X;
double P;
double dp[2][(1 << MAX_M) + 1];
void solve()
{
int n = 1 << M;
double *prv = dp[0], *nxt = dp[1];
memset(prv, 0, sizeof (double ) * (n + 1));
prv[n] = 1.0;
for (int r = 0; r < M; r++) //枚举轮
{
for (int i = n; i >= 0; i--)
{
int jub = min(i, n - i);
double t = 0.0;
for (int j = 0; j <= jub; j++)
{
t = max(t, P * prv[i + j] + (1 - P) * prv[i - j]);
}
nxt[i] = t;
}
swap(prv, nxt);
}
int i = (ll) X * n / (1e6);
// cout << fixed << setprecision(6) << prv[i] << endl;
printf("%.6f\n", prv[i]);
}
int main()
{
freopen("E:\\c2.txt", "r", stdin);
freopen("E:\\out2.txt", "w", stdout);
int k;
// cin >> k;
scanf("%d",&k);
for (int kase = 1; kase <= k; kase++)
{
// cin >> M >> P >> X;
scanf("%d%lf%d", &M, &P, &X);
printf("Case #%d: ", kase);
// cout << "Case #" << kase << ": ";
solve();
}
fclose(stdin);
fclose(stdout);
return 0;
}