Check the results of CsCode Programming Contest 3, click here.
Contest pattern: 2 Programming Questions and 5 MCQ’s.
Contest was organised by @Computer_Science_Engineering .
Answers of MCQ questions
1)
What will be the output of above C program?
Choices
- 123
- 1
- 321123 Correct answer
- 32123
2)
Above code defines which concept?
Choices
- Iteration
- Simple function call
- Recursion Correct answer
- Conditional Statement
3)
What is the time complexity of code below?
Choices
- O(n)
- O(200)
- O(n-200)
- O(1) Correct answer
Check this post for explanation
4)
Tailed recursive functions are slower than non-tailed recursive functions.
Above statement is correct or incorrect?
Choices
- Correct
- Incorrect Correct answer
Check following post for explanation.
5)
Choose the correct output
Choices
- Error Correct answer
- 4
- 3
- 2
Hint:
Solutions of Programming Questions
1) Perfect Number
Problem statement
Write a program to check whether the given number is a perfect number or not. The program should print 1 if the number is a perfect number, else it should print 0.
Hint: Perfect number is a positive whole number that is equal to the sum of its proper divisors.
The first perfect number is 6 as the sum of its proper positive divisors, 1,2 and 3 is 6. Other perfect numbers are 28, 496 and so on.
Input Format:
First line of input contains T test cases.
For each test case,
a positive integer on new line
Output:
Display 1 if number is perfect else 0 for each numbes on new line.
Constraints:
1 <= T <= 1000
1<= Number <= 100000Solution:
def check_perfect_number(number):
if number == 1:
return False
sum_ = 1
for i in range(2, number//2+2):
if number%i == 0:
sum_ += i
if sum_ == number:
return True
return False
for _ in range(int(input())):
number = int(input())
if check_perfect_number(number):
print(1)
else:
print(0)
def isPerfect(n):
s = 1
i = 2
while i*i <= n:
if(n % i ==0):
if n//i == i:
s = s+i
else:
s = s + i + n//i
i += 1
return (1 if s == n and n!=1 else 0)
for i in range(int(input())):
n = int(input())
if isPerfect(n):
print(1)
else:
print(0)
// By kamesh upmanyu
#include<bits/stdc++.h>
using namespace std;
bool check(int n)
{ if(n==1||n==0)
return 0;
int res=0;
for(int i=2;i<=sqrt(n);i++)
{
if(n%i==0)
{
if(i==(n/i))
res+=i;
else
res+=(i+(n/i));
}
}
if(res+1==n)
return 1;
return 0;
}
int main()
{
int t,n;
cin>>t;
while(t--)
{
cin>>n;
cout<<check(n)<<"\n";
}
}
2) Special Sequence – Count Numbers
Problem statement
The following sequence is formed using words and numbers:
- First number is 1
- In the first number, there was one 1 so the second number is 11.
- In the second number, there was two 1s, so the third number is 21
- Like that the fourth number is 1211 (one 2 and one 1)
- The next number is 111221 (one 1, one 2, two 1’s).
This sequence can continue infinitely. Given an integer, q[i], determine the sum of digits of the value in the sequence at that position. For example, position 4 contains the value 1211. The sum of those digits is 1+2+1+1 = 5.
Constraints:
1 <= n <= 1000
1 <= q[i] <= 54Input Format:
The first line contains an integer n (total number of queries).
Next n lines contains an integer q[i].
Output Format:
n lines denoting sum of digits of sequence at q[i]
Check the sample test case solution for more information about output.
Sample Input:
2
2
4Sample Output:
2
5Explanation:
The sequence is 1, 11, 21, 1211, 111221 and so on.
The sum of digits of element at 2nd(11) and 4th(1211) position in the sequence are 2, 5.
Solution:
sumOfDigits = [0 for _ in range(56)]
sumOfDigits[1] = 1
firstNumber = '11'
for i in range(1, 55):
sum_ = 0
for no in firstNumber:
sum_ += int(no)
sumOfDigits[i+1] = sum_
nextNumber = ''
j=1
count = 1
for j in range(1, len(firstNumber)):
if firstNumber[j] == firstNumber[j-1]:
count += 1
else:
nextNumber += str(count)+firstNumber[j-1]
count = 1
nextNumber += str(count)+firstNumber[-1]
firstNumber = nextNumber
#input
for _ in range(int(input())):
print(sumOfDigits[int(input())])
//By kamesh upmanyu
#include<bits/stdc++.h>
using namespace std;
string v[55];;
void create()
{
v[1]=to_string(1);
for(int i=2;i<=54;i++)
{
string str=v[i-1];
string res="";
for(int j=0;j<str.length();j++)
{
int c=1;
while(str[j]==str[j+1])
{ c++;j++;}
res+=to_string(c)+str[j];
}
v[i]=res;
}
}
int main()
{ create();
int t,n;
cin>>t;
while(t--){
int s=0;
cin>>n;
string str=v[n];
for(int i=0;i<str.length();i++)
{
s+=str[i]-'0';
}
cout<<s<<"\n";
}
}
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