To make the solution simpler:

we notice that the ratio must equal K, that means the sum of tastes must be of the form X.K and the sum

of the calories must equal X, so ratio = X.K / X = K , so if we multiplied all calories values by K

then the ratio must be X.K / X.K which equals 1.

and to get the ratio 1 means the nominator equals the denominator which means nominator-denominator=0.

let dp[i][j] mean the maximum value(of the best subset) of tastes we can get from the fruits

1,2,...i where j is the difference between the nominator and the denominator of ratio of the chosen

subset. and the transitions are :

1- dp[i+1][j] = max(dp[i+1][j] , dp[i][j]) when we don't pick the item (i+1)

2- dp[i+1][j+diff] = max(dp[i+1][j+diff] , dp[i][j] + a[i+1]) where diff = a[i+1] — b[i+1] this transition when we pick the item (i+1).

so the result is dp[n][0], but because j might be negative so we make shift to the whole range of the

second dimension of the array dp by the maximum value |-10000| so the result will be dp[n][10000]