SRM 626 D1L1: FixedDiceGameDiv1，贝叶斯公式，dp

Posted 2020-09-14 blfbuaa

题目：http://community.topcoder.com/stat?c=problem_statement&pm=13239&rd=15859

用到了概率论中的贝叶斯公式，而贝叶斯公式中须要用到的概率须要用dp方法求解。

代码：

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair

/*************** Program Begin **********************/
const int MAX_SCORE = 50 * 50;
class FixedDiceGameDiv1 {
public:
	double dp1[MAX_SCORE + 1], dp2[MAX_SCORE + 1];

	// dp[i]: roll a b-sieded dice 最后总得分为i的概率
	void calc(int a, int b, double dp[])
	{
		for (int i = 0; i < MAX_SCORE + 1; i++) {
			dp[i] = 0.0;
		}
		dp[0] = 1.0;
		for (int i = 0; i < a; i++) {
			for (int j = a * b; j >= 0; j--) {
				if (dp[j] == 0) {
					continue;
				}
				for (int k = 1; k <= b; k++) {
					dp[j + k] += dp[j] / b;
				}
				dp[j] = 0;
			}
		}
	}
	double getExpectation(int a, int b, int c, int d) {
		double res = 0.0;

		calc(a, b, dp1);
		calc(c, d, dp2);

		// 贝叶斯公式
		double up = 0, down = 0;
		for (int i = a; i <= a * b; i++) {
			for (int j = 0; j < i; j++) {
				up += dp1[i] * dp2[j] * i;
				down += dp1[i] * dp2[j];
			}
		}

		if (down == 0) {
			return -1;
		}
		res = up / down;

		return res;
	}

};

/************** Program End ************************/