2 and 3
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2

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D
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Self unit employee

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RSPQ

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CBEDA
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Mostly, it is students from English mediun schools who appear in this competition

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Rs:- 1200
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B
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2:1
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33/50

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24
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How many pairs of letters are there in the word "DYNASTY" which has as many letters between them (from both the sides) as in the alphabets from A to Z?

A) 2

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In MS Word, how can a document be split into multiple parts such that each part can have different formats and layouts

Ans)By creating sections

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Which of the statement(s) given below is/are correct?


1)Routers can be static or dynamic
2)Routers connect two or more separate networks,

Ans) both 1 and 2

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X(1), (2)

int count;

// Line 1



Antint i) { count 1)



int getcount () {return count)

// Line 5



public class Book


public static void main(String[] args)


System.out.println(AB.X.count + AB.Y.count); // Line 10


Ans)Compile time error at "// Line 10"

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class TreeNode:
    def init(self, val):
        self.val = val
        self.children = []

def build_tree(units, relationships):
    nodes = {i: TreeNode(units[i]) for i in range(len(units))}
    for rel in relationships:
        parent, child = rel
        nodes[parent].children.append(nodes[child])
    return nodes[0] 

def max_electricity_per_floor(root):
    max_electricity = 0
    floor_electricity = {}
   
    def dfs(node, level):
        nonlocal max_electricity
        if level not in floor_electricity:
            floor_electricity[level] = 0
        floor_electricity[level] += node.val
        max_electricity = max(max_electricity, floor_electricity[level])
        for child in node.children:
            dfs(child, level + 1)
   
    dfs(root, 0)
    return max_electricity


num = int(input())
units = list(map(int, input().split()))
numRel, memConnect = map(int, input().split())
relationships = [tuple(map(int, input().split())) for _ in range(numRel)]


root = build_tree(units, relationships)


result = max_electricity_per_floor(root)
print(result)
Apartment one


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int main() {
ll N;
cin >> N;

vector<ll> w(N);
for(ll i = 0; i < N; ++i) {
cin >> w[i];
}

vector<ll> s(N, 15);
for(ll i = N-2; i >= 0; --i) {
for(ll j = i+1; j < N; ++j) {
if(w[i] < w[j]) {
s[i] = 10;
for(ll k = j+1; k < N; ++k) {
if(w[j] > w[k]) {
s[i] = 5;
break;
}
}
break;
}
}
}

ll g = 0;
for(ll sc : s) {
g += sc;
}

cout << g << endl;

return 0;
}


chair manufacture

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