Codeforces Round #643 (Div. 2) : Young Explorers :solution in python

B. Young Explorers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Young wilderness explorers set off to their first expedition led by senior explorer Russell. Explorers went into a forest, set up a camp and decided to split into groups to explore as much interesting locations as possible. Russell was trying to form groups, but ran into some difficulties...

Most of the young explorers are inexperienced, and sending them alone would be a mistake. Even Russell himself became senior explorer not long ago. Each of young explorers has a positive integer parameter ei

 — his inexperience. Russell decided that an explorer with inexperience e can only join the group of e

or more people.

Now Russell needs to figure out how many groups he can organize. It's not necessary to include every explorer in one of the groups: some can stay in the camp. Russell is worried about this expedition, so he asked you to help him.

Input

The first line contains the number of independent test cases T

(1≤T≤2⋅105). Next 2T

lines contain description of test cases.

The first line of description of each test case contains the number of young explorers N

(1≤N≤2⋅105

).

The second line contains N

integers e1,e2,…,eN (1≤ei≤N), where ei is the inexperience of the i

-th explorer.

It's guaranteed that sum of all N

doesn't exceed 3⋅105

.

Output

Print T

numbers, each number on a separate line.

In i

-th line print the maximum number of groups Russell can form in i

-th test case.

Example

Input

Copy

2
3
1 1 1
5
2 3 1 2 2

Output

Copy

3
2

Note

In the first example we can organize three groups. There will be only one explorer in each group. It's correct because inexperience of each explorer equals to 1

, so it's not less than the size of his group.

In the second example we can organize two groups. Explorers with inexperience 1

, 2 and 3 will form the first group, and the other two explorers with inexperience equal to 2

will form the second group.

This solution is not unique. For example, we can form the first group using the three explorers with inexperience equal to 2

, and the second group using only one explorer with inexperience equal to 1. In this case the young explorer with inexperience equal to 3 will not be included in any group.

Solution in Python :

t = int(input())
ar = []
for i in range(t) :
    n = int(input())
    e = [int(i) for i in input().split()]
    e.sort()
    mul = 0 ; count = 0 ; count1 = 1 ; sumif = 0 ;sumelse = 0
    for i in e:
        if i == 1 :
            count += 1
        else :
            if mul <count1**2 :
                if count1 == 1:
                    mul = 1
                mul *= i
                count1 += 1
                sumif += 1
            else :
                count1 = 1
                mul = 0
                count += 1
                sumelse += 1
    if sumif <sumelse :
        count -= 1
    elif sumif >sumelse :
        count += 1
    ar.append(count)
for i in ar :
    print(i)

Note : some modification required !! Hackerearth Byteland Problem : Solution in python

 PROBLEM STATEMENT

Points: 50

In Byteland they have a very strange monetary system. Each Bytelandian gold coin has an integer number written on it. A coin n can be exchanged in a bank into three coins: n/2, n/3 and n/4. But these numbers are all rounded down (the banks have to make a profit).

You can also sell Bytelandian coins for American dollars. The exchange rate is 1:1. But you can not buy Bytelandian coins. You have one gold coin. What is the maximum amount of American dollars you can get for it?

Input The input will contain several test cases (not more than 10). Each testcase is a single line with a number n, 0 <= n <= 1 000 000 000. It is the number written on your coin.

Output For each test case output a single line, containing the maximum amount of American dollars you can make.

Explanation You can change 12 into 6, 4 and 3, and then change these into $6+$4+$3 = $13. If you try changing the coin 2 into 3 smaller coins, you will get 1, 0 and 0, and later you can get no more than $1 out of them. It is better just to change the 2 coin directly into $2.

SAMPLE INPUT

 

12
2

SAMPLE OUTPUT

 

13
2

Time Limit: 9.0 sec(s) for all input files combined.

Solution in python :

list1 = []

while True :

    try :

        n = input()

        if n :

            count = 0

            n = int(n)

            count+=int(n/2)+int(n/3)+int(n/4)

            if n>count :

               count = n

            list1.append(count)

    except EOFError :

           break

for i in list1 :

    print(i) 

Hackerearth : checking horizontal and vertical symmetric matrix : Solution in python

PROBLEM STATEMENT

Points: 30

You are given a square matrix of size n. Rows are indexed 1 to n from top to bottom and columns are indexed 1 to n form left to right. Matrix consists of only '*' and '.'. You need to check whether matrix is symmetric or not. if it is, check it is symmetric about vertical axis or horizontal axis or both.

A matrix is said to be symmetric about horizontal axis if $1st$ row is identical to $nth$ row, $2nd$ is identical to $(n-1)th$ row and so on...

A matrix is said to be symmetric about vertical axis if $1st$ column is identical to nth column, $2nd$ identical to $(n-1)th$ and so on for all columns.

INPUT :

First line contains t,the number of test cases. First line of each test case contains n the size of matrix. Each of next n lines contain n characters.

OUTPUT:

Output t lines, answer for each test case. Print "HORIZONTAL" if symmetric about horizontal axis. Print "VERTICAL" if symmetric about vertical axis. Print "BOTH" if symmetric about both axes. print "NO" if it is not symmetric.

Constraints :

$1<t\le 500$
$1<n<50$

SAMPLE INPUT

 

3
4
*.*.
.*.*
*.*.
.*.*
3
.*.
*.*
.*.
3
..*
**.
..*

SAMPLE OUTPUT

 

NO
BOTH
HORIZONTAL

Solution in python:

list2 = []

t = int(input())

for i in range(t) :

    list1 = []

    count = 0 

    count2 = 0

    n  = int(input())

    for j in range(0,n) :

        list1.append(input().split([0])

    for k in range(0,n) :

        if list1[0+k] != list1[n-1-k] :

            count = 0

            break 

        else : 

            count=1

    for j in range(n) :

        for k in range(n) :

                if list1[0+k][0+j] != list1[0+k][n-1-j] :

                    count2 = 0

                    break 

                else : 

                    count2 = 1

    if count==0 and count2==0 :

        list2.append('NO')

    elif count==0 and count2 :

        list2.append('VERTICAL')

    elif count and count2 == 0 :

         list2.append('HORIZONTAL')

    else :

         list2.append('BOTH')

for i in list2 :

    print(i)