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// 808. Soup Servings
#include <stdio.h>
#include <stdlib.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.
*/
double process(int a, int b, double **record) {
if (a <= 0 && b <= 0)
return 0.5;
if (a <= 0)
return 1.0;
if (b <= 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 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;
}
int main() {
printf("%f\n", soupServings(50)); // expect: 0.625000
printf("%f\n", soupServings(100)); // expect: 0.71875
}
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