// 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 }