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// 808. Soup Servings
#include "leetcode.h"
/*
* there are two ops of soup: op a, and op b. initially, we have 'n' ml of
* each op 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 ops 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.
*/
double probability(int a, int b) {
if (a <= 0 && b > 0)
return 1;
if (a <= 0 && b <= 0)
return 0.5;
if (b <= 0 && a > 0)
return 0;
double op1 = probability(a - 100, b);
double op2 = probability(a - 75, b - 25);
double op3 = probability(a - 50, b - 50);
double op4 = probability(a - 25, b - 75);
return 0.25 * (op1 + op2 + op3 + op4);
}
double soupServings(int n) {
if (n >= 5000)
return 1;
return probability(n, n);
}
int main() {
printf("%f\n", soupServings(50)); // expect: 0.625000
printf("%f\n", soupServings(100)); // expect: 0.71875
}
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