2008 APAC local onsites C Millionaire （动态规划,离散化思想）

Problem

You have been invited to the popular TV show "Would you like to be a millionaire?". Of course you would!

The rules of the show are simple:

Before the game starts, the host spins a wheel of fortune to determine P, the probability of winning each bet.

You start out with some money: X dollars.

There are M rounds of betting. In each round, you can bet any part of your current money, including none of it or all of it. The amount is not limited to whole dollars or whole cents.

If you win the bet, your total amount of money increases by the amount you bet. Otherwise, your amount of money decreases by the amount you bet.

After all the rounds of betting are done, you get to keep your winnings (this time the amount is rounded down to whole dollars) only if you have accumulated $1000000 or more. Otherwise you get nothing.

Given M, P and X, determine your probability of winning at least $1000000 if you play optimally (i.e. you play so that you maximize your chances of becoming a millionaire).

Input

The first line of input gives the number of cases, N.

Each of the following N lines has the format "M P X", where:

M is an integer, the number of rounds of betting.

P is a real number, the probability of winning each round.

X is an integer, the starting number of dollars.

Output

For each test case, output one line containing "Case #X: Y", where:

X is the test case number, beginning at 1.

Y is the probability of becoming a millionaire, between 0 and 1.

Answers with a relative or absolute error of at most 10-6 will be considered correct.

Limits

1 ≤ N ≤ 100

0 ≤ P ≤ 1.0, there will be at most 6 digits after the decimal point.

1 ≤ X ≤ 1000000

Small dataset

1 ≤ M ≤ 5

Large dataset

1 ≤ M ≤ 15

Sample

In the first case, the only way to reach $1000000 is to bet everything in the single round.

In the second case, you can play so that you can still reach $1000000 even if you lose a bet. Here's one way to do it:

You have $600000 on the first round. Bet $150000.

If you lose the first round, you have $450000 left. Bet $100000.

If you lose the first round and win the second round, you have $550000 left. Bet $450000.

If you win the first round, you have $750000 left. Bet $250000.

If you win the first round and lose the second round, you have $500000 left. Bet $500000.

题意：

最开始你有x元钱，要进行M轮赌博。每一轮赢的概率为P，你可以选择赌与不赌，如果赌也可以将所持的任意一部分钱作为赌注（可以是整数，也可以是小数）。如果赢了，赌注将翻倍；输了赌注则没了。在M轮赌博结束后，如果你持有的钱在100万元以上，就可以把这些钱带回家。问：当你采取最优策略时，获得100万元以上的钱并带回家的概率是多少。

分析：

由于每一轮的赌注是任意的，不一定为整数，因而有无限种可能，所以即便想穷竭搜索也无从着手。但如果能化连续为离散，那么可能便也是有限的了。具体如下：假设前M-1轮的赌博后，还持有x'元。对于最后一轮，考虑的情况有3种。如果x' >= 100万，则没有必要再赌了即最后一轮赢的概率为0；如果50<= x' < 100万，只要参与赌博并且赌注 >= 50万则有赢的概率为P；如果x' < 50万，那么无论是否参与最后一轮的赌博，压的赌注是多少赢的概率必为0。我们不妨看一下倒数第二轮与最后一轮的关系，设在倒数第二轮时持有的钱为x。如果x >= 100万，赢的概率为1；如果x < 25万，即便最后两轮赌博都赢了总钱数必小于100万，所以赢的概率为0；否则，只要选择参与至少一轮赌博并且赌注至少25万则有赢得概率。假设倒数第二轮的赌注为y(y = 0 或 y >= 25万)，则最后一轮持有的钱x' = (x + y)或x' = (x - y)。而倒数第二轮考虑的情况具体可以分为5种。综上，当参与M轮赌博时所需考虑的情况总共有2^m + 1种，可以通过dp解决。定义一个二维dp数组，dp[i][j] := 参与第i轮赌博，持有的钱所在模块为j并且采取最优策略时赢的概率。初始化：dp[n][1 << m] = 1，状态转移方程dp[i][j] = max(P * dp[i + 1][j + k] + (1 - P) * dp[i + 1][j - k] / 0 <= k <= min(j, n - j) )。时间复杂度O(m*2^2m)。

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;
typedef long long int LL;
int M , X  ;
double P;
double dp[2][(1 << 15) + 1];
void solve()
{
    int n = 1 << M;
    double *pre = dp[0] , *nxt = dp[1];
    memset(pre , 0 , sizeof(double) * (n + 1));
    ///memset(pre , 0 , sizeof(pre)); 这样初始化是不行的，因为pre为一个double型的指针，不是整个数组。
    pre[n] = 1.0;///因：模块n对应的资金>= 100万
    for(int r = 0 ; r < M; r++)///枚举第几轮
    {
        for(int i = 0 ; i <= n ; i++)///枚举当前是哪种状态
        {
            int step = min(i , n - i);///如果step大于n / 2 ， 等会儿转移的时候可能会超过n
            double t = 0.0;
            for(int j = 0 ; j <= step ; j++)///枚举当前的所有可能走法
            {
                t = max(t , P * pre[i + j] + (1 - P) * pre[i - j]);///求出期望的最大值
            }
            nxt[i] = t;
        }
        swap(pre , nxt);///交换两个数组的值进行滚动
    }
    int i = (LL)X * n / 1000000;///找到X对应的是第几块
//    for(int i = 0 ; i <= n ; i++)cout << '*' << pre[i] << endl;
    printf("%.6lf\n" , pre[i]);
}
int main()
{
    cin >> M >> P >> X;
    solve();
    return 0;
}