Weekly Contest 78-------->808. Soup Servings-优快云博客

探讨了在两种汤（A和B）之间进行不同操作的概率问题，目标是在特定条件下预测哪种汤会先被消耗完。通过使用动态规划方法，在C++、Java和Python中实现了算法解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There are two types of soup: type A and type B. Initially we have N ml of each type of soup. There are four kinds of operations:

Serve 100 ml of soup A and 0 ml of soup B
Serve 75 ml of soup A and 25 ml of soup B
Serve 50 ml of soup A and 50 ml of soup B
Serve 25 ml of soup A and 75 ml of soup B

When we serve some soup, we give it to someone and we no longer have it. Each turn, we will choose from the four operations with equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as we can. We stop once we no longer have some quantity of both types of soup.

Note that we do not have the operation where all 100 ml's of soup B are used first.

Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time.

Example:
Input: N = 50
Output: 0.625
Explanation: 
If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.

Notes:

0 <= N <= 10^9.
Answers within 10^-6 of the true value will be accepted as correct.

Approach #1: C++. Using DP.

class Solution {
private:
    double memo[200][200];
    
public:
    double soupServings(int N) {
        return N > 4800 ? 1.0 : solve((N+24) / 25, (N + 24) / 25);
    }
private:
    double solve(int a, int b) {
        if (a <= 0 && b <= 0) return 0.5;
        if (a <= 0) return 1.0;
        if (b <= 0) return 0.0;
        if (memo[a][b] > 0) return memo[a][b];
        memo[a][b] = 0.25 * (solve(a-4, b) + solve(a-3, b-1) + solve(a-2, b-2) + solve(a-1, b-3));
        return memo[a][b];
    }
};

Approach #2: Java.

class Solution {
    private double[][] memo = new double[200][200];
    
    public double soupServings(int N) {
        return N > 4800 ? 1.0 : solve((N + 24) / 25, (N + 24) / 25);
    }
    
    private double solve(int a, int b) {
        if (a <= 0 && b <= 0) return 0.5;
        if (a <= 0) return 1.0;
        if (b <= 0) return 0.0;
        if (memo[a][b] > 0) return memo[a][b];
        memo[a][b] = 0.25 * (solve(a-4, b) + solve(a-3, b-1) + solve(a-2, b-2) + solve(a-1, b-3));
        return memo[a][b];
    }
}

Approach #3: Python.

class Solution(object):
    memo = {}
    def soupServings(self, N):
        if N > 4800: return 1
        def f(a, b):
            if (a, b) in self.memo: return self.memo[a, b]
            if a <= 0 and b <= 0: return 0.5
            if a <= 0: return 1
            if b <= 0: return 0
            self.memo[(a, b)] = 0.25 * (f(a - 4, b) + f(a - 3, b - 1) + f(a - 2, b - 2) + f(a - 1, b - 3))
            return self.memo[(a, b)]
        N = math.ceil(N / 25.0)
        return f(N, N)

转载于:https://www.cnblogs.com/ruruozhenhao/p/9974242.html