Berland Fair
Contest: Educational Codeforces Round 53 (Rated for Div. 2)
Difficulty rating: 1700
Tags: binary search, brute force, data structures, greedy

XXI Berland Annual Fair is coming really soon! Traditionally fair consists of $$$n$$$ booths, arranged in a circle. The booths are numbered $$$1$$$ through $$$n$$$ clockwise with $$$n$$$ being adjacent to $$$1$$$. The $$$i$$$-th booths sells some candies for the price of $$$a_i$$$ burles per item. Each booth has an unlimited supply of candies.

Polycarp has decided to spend at most $$$T$$$ burles at the fair. However, he has some plan in mind for his path across the booths:

- at first, he visits booth number $$$1$$$;
- if he has enough burles to buy exactly one candy from the current booth, then he buys it immediately;
- then he proceeds to the next booth in the clockwise order (regardless of if he bought a candy or not).

Polycarp's money is finite, thus the process will end once he can no longer buy candy at any booth.

Calculate the number of candies Polycarp will buy.

Input format:
The first line contains two integers $$$n$$$ and $$$T$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le T \le 10^{18}$$$) — the number of booths at the fair and the initial amount of burles Polycarp has.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the price of the single candy at booth number $$$i$$$.

Output format:
Print a single integer — the total number of candies Polycarp will buy.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
Let's consider the first example. Here are Polycarp's moves until he runs out of money:

1. Booth $$$1$$$, buys candy for $$$5$$$, $$$T = 33$$$;
2. Booth $$$2$$$, buys candy for $$$2$$$, $$$T = 31$$$;
3. Booth $$$3$$$, buys candy for $$$5$$$, $$$T = 26$$$;
4. Booth $$$1$$$, buys candy for $$$5$$$, $$$T = 21$$$;
5. Booth $$$2$$$, buys candy for $$$2$$$, $$$T = 19$$$;
6. Booth $$$3$$$, buys candy for $$$5$$$, $$$T = 14$$$;
7. Booth $$$1$$$, buys candy for $$$5$$$, $$$T = 9$$$;
8. Booth $$$2$$$, buys candy for $$$2$$$, $$$T = 7$$$;
9. Booth $$$3$$$, buys candy for $$$5$$$, $$$T = 2$$$;
10. Booth $$$1$$$, buys no candy, not enough money;
11. Booth $$$2$$$, buys candy for $$$2$$$, $$$T = 0$$$.

No candy can be bought later. The total number of candies bought is $$$10$$$.

In the second example he has $$$1$$$ burle left at the end of his path, no candy can be bought with this amount.

Example input:
3 38
5 2 5

Example output:
10

Example input:
5 21
2 4 100 2 6

Example output:
6

Editorial:
Let's code the following process. Go one circle across the booths, calculate the total cost $$$C$$$ of sweets bought and the number $$$S$$$ of sweets bought. Now you can decrease you money down to $$$T = T~mod~C$$$ and add $$$S \cdot (T~div~C)$$$ to answer. It represents that you went maximum number of such circles. The later circles will have smaller cost. Let's continue this process until $$$T$$$ becomes smaller than the minimum priced sweet.
The number of operations made is $$$O(\log T)$$$. Let $$$T_{cur}$$$ be the amount of money before some operation, $$$C_{cur}$$$ be the total cost of sweets bought on that operation and $$$T_{new} = T_{cur}~mod~C_{cur}$$$. $$$T_{new}$$$ is actually smaller than $$$C_{cur}$$$ (that's how modulo works) and smaller than $$$T_{cur} - C_{cur}$$$ (that's also how modulo works). And these inequalities imply that $$$T_{new} < \frac{T_{cur}}{2}$$$. That leads to about $$$O(\log T)$$$ steps to reach the minimal price.
Overall complexity: $$$O(n \log T)$$$.