GCJ 2008 APAC local onsites C Millionaire

/*
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
", 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;
}