Copy-paste (Codeforces 1417A)
Tags: Greedy, math

 — Hey folks, how do you like this problem?— That'll do it. BThero is a powerful magician. He has got $$$n$$$ piles of candies, the $$$i$$$-th pile initially contains $$$a_i$$$ candies. BThero can cast a copy-paste spell as follows:   He chooses two piles $$$(i, j)$$$ such that $$$1 \le i, j \le n$$$ and $$$i \ne j$$$.  All candies from pile $$$i$$$ are copied into pile $$$j$$$. Formally, the operation $$$a_j := a_j + a_i$$$ is performed. BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than $$$k$$$ candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?

Input: The first line contains one integer $$$T$$$ ($$$1 \le T \le 500$$$) — the number of test cases.Each test case consists of two lines:   the first line contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le n \le 1000$$$, $$$2 \le k \le 10^4$$$);  the second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i \le k$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$1000$$$, and the sum of $$$k$$$ over all test cases does not exceed $$$10^4$$$.

Output: For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.

Note: In the first test case we get either $$$a = [1, 2]$$$ or $$$a = [2, 1]$$$ after casting the spell for the first time, and it is impossible to cast it again.

Sample Inputs:
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2

Sample Outputs:
1
5
4

Solution (Python):
""""""

def slv():
    (n, k) = map(int, input().split())
    a = list(map(int, input().split()))
    cnt = 0
    a.sort()
    for i in range(1, n):
        cnt += (k - a[i]) // a[0]
    print(cnt)
if __name__ == '__main__':
    t = 1
    t = int(input())
    while t > 0:
        slv()
        t -= 1

Solution (Python):
for _ in range(int(input())):
    (n, k) = list(map(int, input().split(' ')))
    a = sorted(list(map(int, input().split(' '))))
    s = 0
    for i in range(1, n):
        s += (k - a[i]) // a[0]
    print(s)

Solution (Python):
for t in range(int(input())):
    (n, k) = list(map(int, str.split(input())))
    a = sorted(map(int, str.split(input())))
    rslt = 0
    for i in range(1, n):
        rslt += (k - a[i]) // a[0]
    print(rslt)