Increase and Copy (Codeforces 1426C)
Tags: Math, binary search

Initially, you have the array $$$a$$$ consisting of one element $$$1$$$ ($$$a = [1]$$$).In one move, you can do one of the following things:  Increase some (single) element of $$$a$$$ by $$$1$$$ (choose some $$$i$$$ from $$$1$$$ to the current length of $$$a$$$ and increase $$$a_i$$$ by one);  Append the copy of some (single) element of $$$a$$$ to the end of the array (choose some $$$i$$$ from $$$1$$$ to the current length of $$$a$$$ and append $$$a_i$$$ to the end of the array). For example, consider the sequence of five moves:  You take the first element $$$a_1$$$, append its copy to the end of the array and get $$$a = [1, 1]$$$.  You take the first element $$$a_1$$$, increase it by $$$1$$$ and get $$$a = [2, 1]$$$.  You take the second element $$$a_2$$$, append its copy to the end of the array and get $$$a = [2, 1, 1]$$$.  You take the first element $$$a_1$$$, append its copy to the end of the array and get $$$a = [2, 1, 1, 2]$$$.  You take the fourth element $$$a_4$$$, increase it by $$$1$$$ and get $$$a = [2, 1, 1, 3]$$$. Your task is to find the minimum number of moves required to obtain the array with the sum at least $$$n$$$.You have to answer $$$t$$$ independent test cases.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then $$$t$$$ test cases follow.The only line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 10^9$$$) — the lower bound on the sum of the array.

Output: For each test case, print the answer: the minimum number of moves required to obtain the array with the sum at least $$$n$$$.


Sample Inputs:
5
1
5
42
1337
1000000000

Sample Outputs:
0
3
11
72
63244

Solution (Python):
import math
for _ in range(int(input())):
    n = int(input())
    print(math.ceil((n ** 0.5 - 1) * 2))

Solution (Python):
from math import *
for s in [*open(0)][1:]:
    print(ceil(2 * sqrt(int(s))) - 2)

Solution (Python):
from math import *
for s in [*open(0)][1:]:
    print(ceil(2 * sqrt(int(s))) - 2)