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-// 808. Soup Servings
-#include "leetcode.h"
-
-/*
- * 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:
- * 1. serve 100 ml of soup a and 0ml of soup b
- * 2. serve 75ml of soup a and 25ml of soup b
- * 3. serve 50ml of soup a and 50ml of soup b
- * 4. serve 25ml of soup a and 75ml 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 an equal probability
- * of '0.25'. if the remaining volume of soup is not enough to complete the
- * operation, we will serve as much as possible. we stop once we no longer have
- * some quantity of both types of soup. note that we do not have an 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. answers within '10^-5' of the actual answer will be accepted.
- */
-
-class Solution {
- unordered_map<int, unordered_map<int, double>> store;
- double helper(int a, int b) {
- if (!a && b)
- return 1;
- else if (!a && !b)
- return 0.5;
- else if (a && !b)
- return 0;
- if (store[a][b])
- return store[a][b];
- double prob = 0;
- prob += 0.25 * helper(a - min(a, 100), b);
- prob += 0.25 * helper(a - min(a, 75), b - min(b, 25));
- prob += 0.25 * helper(a - min(a, 50), b - min(b, 50));
- prob += 0.25 * helper(a - min(a, 25), b - min(b, 75));
- store[a][b] = prob;
- return prob;
- }
-
-public:
- double soupServings(int n) {
- if (n >= 5000)
- return 1.0;
- else
- return helper(n, n);
- }
-};
-
-int main() {
- Solution obj;
- printf("%f\n", obj.soupServings(50)); // expect: 0.62500
- printf("%f\n", obj.soupServings(100)); // expect: 0.71875
-}