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author | Jack Sangdahl <[email protected]> | 2025-08-08 11:43:46 +0200 |
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committer | Jack Sangdahl <[email protected]> | 2025-08-08 11:43:46 +0200 |
commit | f878bb71563de831927bdf52b44e2dcc0ab4c577 (patch) | |
tree | 1db19de3971c90fbfb230d884aa3293187d49f9b | |
parent | 3b6f252059951a3a0927dfc15993ce1382f70e76 (diff) | |
download | leetcode-f878bb71563de831927bdf52b44e2dcc0ab4c577.tar.gz leetcode-f878bb71563de831927bdf52b44e2dcc0ab4c577.zip |
808. soup servings
-rw-r--r-- | soup_servings.c | 39 | ||||
-rw-r--r-- | soup_servings.py | 48 |
2 files changed, 64 insertions, 23 deletions
diff --git a/soup_servings.c b/soup_servings.c index 68a3d9d..6449342 100644 --- a/soup_servings.c +++ b/soup_servings.c @@ -1,10 +1,9 @@ // 808. Soup Servings -#include <stdio.h> -#include <stdlib.h> +#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: + * 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 @@ -13,36 +12,30 @@ * 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 + * 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 process(int a, int b, double **record) { +double probability(int a, int b) { + if (a <= 0 && b > 0) + return 1; if (a <= 0 && b <= 0) return 0.5; - if (a <= 0) - return 1.0; - if (b <= 0) + if (b <= 0 && a > 0) return 0; - if (record[a][b]) - return record[a][b]; - record[a][b] = - 0.25 * (process(a - 4, b, record) + process(a - 3, b - 1, record) + - process(a - 2, b - 2, record) + process(a - 1, b - 3, record)); - return record[a][b]; + 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) { - double **record = malloc(150 * sizeof(double *)); - for (int i = 0; i < 150; i++) - record[i] = calloc(150, sizeof(double)); - double ans = n >= 3600 ? 1.0 : process((n + 24) / 25, (n + 24) / 25, record); - for (int i = 0; i < 150; i++) - free(record[i]); - free(record); - return ans; + if (n >= 5000) + return 1; + return probability(n, n); } int main() { diff --git a/soup_servings.py b/soup_servings.py new file mode 100644 index 0000000..15d7eaf --- /dev/null +++ b/soup_servings.py @@ -0,0 +1,48 @@ +# 808. Soup Servings + +""" +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. +""" + + +class Solution(object): + def probability(self, a, b): + if a <= 0 and b > 0: + return 1 + if a <= 0 and b <= 0: + return 0.5 + if b <= 0 and a > 0: + return 0 + op1 = self.probability(a - 100, b) + op2 = self.probability(a - 75, b - 25) + op3 = self.probability(a - 50, b - 50) + op4 = self.probability(a - 25, b - 75) + return 0.25 * (op1 + op2 + op3 + op4) + + def soupServings(self, n): + """ + :type n: int + :rtype: float + """ + if n > 5000: + return 1.0 + return self.probability(n, n) + + +if __name__ == "__main__": + obj = Solution() + print(obj.soupServings(50)) + print(obj.soupServings(100)) |