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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
}
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