Output #1
10
15
50
For each $i = 1, 2, 3, 4, 5$, let $A_i$ be the number of fruits remaining on the $i$\-th tree. Now, we use the sequence $A = (A_1, A_2, A_3, A_4, A_5)$ to represent the numbers of fruits remaining in the trees.
* On Day $0$, we have $A = (0, 0, 0, 0, 0)$.
* On the morning of Day $1$, each tree bears new fruits, and we have $A = (1, 2, 3, 4, 5)$.
* On the morning of Day $2$, each tree bears new fruits, and we have $A = (2, 4, 6, 8, 10)$.
* On the night of Day $2$, Takahashi performs the $1$\-st harvesting. $4 + 6 = 10$ fruits are collected, and we have $A = (2, 0, 0, 8, 10)$.
* On the morning of Day $3$, each tree bears new fruits, and we have $A = (3, 2, 3, 12, 15)$.
* On the night of Day $3$, Takahashi performs the $2$\-nd harvesting. $3 + 12 = 15$ fruits are collected, and we have $A = (3, 2, 0, 0, 15)$.
* On the morning of Day $4$, each tree bears new fruits, and we have $A = (4, 4, 3, 4, 20)$.
* On the morning of Day $5$, each tree bears new fruits, and we have $A = (5, 6, 6, 8, 25)$.
* On the night of Day $5$, Takahashi performs the $3$\-rd harvesting. $5 + 6 + 6 + 8 + 25 = 50$ fruits are collected, and we have $A = (0, 0, 0, 0, 0)$.