Mu Sigma Pseudo Code Quiz – 1

#include<iostream>

using namespace std;

int main()

{

int x=0;

int n=1;

while (n <= 100)

{

x = x + n;

n = n + 1;

}

cout<<x;

}

 

The given code is in C++

#include<iostream>
using namespace std;
int main()
{
int x =1;
int x1 =0;
int x2 = 0;
int x3 = 1;
while(x<10)
{
x1=x1+x2+x3;
x2=x2+x1+x3;
x3=x3+x2+x1;
cout<<x1<<" ";
cout<<x2<<" ";
cout<<x3<<" ";
x=x+1;
}
return 0;
}

 

#include<iostream>

using namespace std;

int main()

{

int a,b,c;

a=6,b=84;

while(b>0)

{

b=b/2;

a=a+6;

c=a+b;

while(c>40)

{

if(c%2 == 0)

cout<<a;

else

cout<<b;

c=c/10;

}

}

cout<<c;

return 0;

}

 

Lets understand it step by step

initially a=6 , b=84
compiler will enter while loop
1st iteration – b=42 , a=12, c= a+b = 54(compiler will enter 2nd while loop)
(if condition gets true) – it will print a = 12
2nd iteration – b=21, a=18, c= a+b = 39(compiler will not enter 2nd while loop)
3rd iteration – b=10, a=24, c= a+b = 34(compiler will not enter 2nd while loop)
4th iteration – b=5, a=10, c= a+b = 35(compiler will not enter 2nd while loop)
5th iteration – b=2, a=36, c= a+b = 38(compiler will not enter 2nd while loop)
6th iteration – b=1, a=42, c= a+b = 43(compiler will enter 2nd while loop)
(if condition gets false), else will be executed and it will print b – 1
now, c=c/10, that makes c=4
7th iteration – b=0, a=48, c= a+b = 48(compiler will enter 2nd while loop)
(if condition gets true), it will print a – 48
now, c=c/10, that makes c=4
compiler get out of both the while loops
it will print c – that is 4
Hence the output will be
12 1 48 4
We hope, this will solve out your queries

#include<iostream>
using namespace std;
int fun(int i)
{
if((i % 2) != 0)
return i;
else
return fun(fun(i-1));
}
int main()
{
cout<<fun(140);
return 0;
}

The algorithm in this question resembles the algorithm of binary search tree, hence option 2 is the correct option

#include<iostream>
using namespace std;
int main()
{
int a,b;
a=2, b=50;
while(b>0)
{
a = (b%2) + a;
if( a%3==0)
cout<<a;
else
cout<<b-1;
b=b/5;
a=a+1;
}
return 0;
}

#include<iostream>
using namespace std;
int main()
{
int a, b, c, d, e;
a=50 , b=3, c=3, e=0;
while(c>0)
{
d=a % b;
e= e + d + a;
c= c - 1;
}
cout<<e;
return 0;
}

#include<iostream>
using namespace std;
int main()
{
int array1[10] = {2, 3, 56, 34};
int k, a, j, n;
a = 3, n = 4;
for(k=0 ;k<=n-1;k++)
{
array1[n] = array1[0];
for(j=0;j<=n-1;j++)
array1[j] = array1[j+1];
}
for(k=0;k<=n-1;k++)
cout<<array1[k]<<" ";
return 0;
}

[ Note: &-Bitwise AND operator, it takes two numbers as operands and does AND on every bit of two numbers. The result of AND is 1 only if both bits are 1

>> - Bitwise right shift operator, it takes two numbers, right shifts the bits of the first operand, the second operand decides the number of places to shift ]

#include<iostream>
using namespace std;
int main()
{
int j,i,count,num;
j=31, count=0, num=64;
while(num!=0)
{
if((num & 1) == 1)
break;
else
count=count+1;
num = num>>1;
}
cout<<count;
return 0;
}

#include<iostream>
using namespace std;
int main()
{
int a =16;
if((a % 10)== 0)
a=a*2;
else if((a % 5 )== 0)
a=a/5;
else
a=a-1;
cout<<a;
return 0;
}