Solutions for Skillrack Code Test CSE1001 Practice Problems 8
3-Nov-2016 00:00 to 12-Nov-2016 00:00
Total time : 500 mins
Challenges : 5
Question 1 (Identify machines in same local networks)
Numeric addresses for computers on the international network Internet are composed of four parts, separated by periods, of the form xx.yy.zz.mm where xx , yy , zz , and mm are positive integers. Locally, computers are also known by a nicknames.
Sample Data
IP address
111.22.3.44
555.66.7.88
111.22.5.66
0.0.0.0
A pair of computers are said to be in same locality when the first two components of the addresses are same. Given the IP addresses of ‘n’ computers, write an algorithm and the subsequent ‘C’ program to identify the pair of computers from the same locality.
In this example, the message to be displayed will be ‘Machines 1 and 3 are on the same local network’. Index of the machines is printed in the message. Index of the machines start from index 1.
Input Format
Number of computers
IP address of the computer1 as a String
IP address of the computer2 as a String
….
Output Format
For each pair of machines in the same local network, print:
Machines A and B are on the same local network (A and B are index of the machines)
Solution
#include < stdio.h >
#include < string.h >
int main()
{
int n,p,i,j;
char s[20][20];
scanf("%d",&n);
for(i=0;i < n;i++)
{
scanf("%s",s[i]);
p=0;
j=0;
while(p!=2)
{
j++;
if(s[i][j]=='.')
p++;
}
s[i][j]='\0';
}
for(i=0;i < n-1;i++)
for(j=i+1;j < n;j++)
if(strcmp(s[i],s[j])==0)
printf("Machines %d and %d are on the same local network",i+1,j+1);
return(0);
}
Input
INPUT:
n - number of entries
for(i=0;i < n;i++)
{
scanf("%s",s[i]);
p=0;
j=0;
while(p!=2)
{
j++;
if(s[i][j]=='.')
p++;
}
s[i][j]='\0';
}
Output
OUTPUT:
if(strcmp(s[i],s[j])==0)
printf("Machines %d and %d are on the same local network",i+1,j+1);
Processing
for(i=0;i < n-1;i++)
for(j=i+1;j < n;j++)
if(strcmp(s[i],s[j])==0)
Pseudocode
BEGIN
Read n and array s
for(i=0;i < n-1;i++)
for(j=i+1;j < n;j++)
if(strcmp(s[i],s[j])==0)
printf("Machines %d and %d are on the same local network",i+1,j+1);
END
Question 2 (Excess numbers)
Given ‘n’ integers, write an algorithm and the subsequent ‘C’ program to print all Excess numbers in it. A number is said to an ‘Excess number’ if it is greater than the average of the elements in its right. Last number is always considered to be ‘Excess number’. For example, given seven numbers 23, 34, 12, 21, 14, 26, 33, then the numbers 34, 33 are excess numbers.
Input format
First line contains the number of elements, n
Next ‘n’ lines contain the elements
Output format
Print all excess numbers in the collection
Solution
#include < stdio.h >
int main()
{
int n,i,j;
int s=0;
int a[20];
scanf("%d",&n);
for(i=0;i < n;i++)
scanf("%d",&a[i]);
for(i=0;i < n-1;i++)
{
s=0;
for(j=i+1;j < n;j++)
s+=a[j];
s/=(n-i-1);
if(a[i] > s)
printf("%d\n",a[i]);
}
printf("%d",a[n-1]);
return(0);
}
Input
INPUT:
n - number of elements
for(i=0;i < n;i++)
scanf("%d",&a[i]);
Output
OUTPUT:
if(a[i] > s)
printf("%d\n",a[i]);
printf("%d",a[n-1]);
Processing
for(i=0;i < n-1;i++)
{
s=0;
for(j=i+1;j < n;j++)
s+=a[j];
s/=(n-i-1);
}
Pseudocode
BEGIN
Read n and array a
for(i=0;i < n-1;i++)
{
s=0;
for(j=i+1;j s)
printf("%d\n",a[i]);
}
printf("%d",a[n-1]);
END
Question 3 (Singular to plural)
A Software is developed for kids to give plural of a given word. Write an algorithm and the Subsequent ‘C’ program to read a noun and print their plurals on the basis of the following rules:
a. If noun ends in “y”, remove the “y” and add “ies”.
b. If noun ends in “s”, “ch”, or “sh”, add “es”.
c. In all other cases, just add “s”.
Print each noun and its plural in lowercase letters with separated by a space.
Input format
Noun – A String
Output format
Plural of the noun
Solution
#include < string.h >
#include < stdio.h >
int main()
{
char n[20];
int i,j,l;
scanf("%s",&n);
l=strlen(n);
if(n[l-1]=='y')
{
n[l-1]='i';
n[l]='e';
n[l+1]='s';
n[l+2]='\0';
}
else if(n[l-1]=='s'||(n[l-2]=='c'&&n[l-1]=='h')||(n[l-2]=='s'&&n[l-1]=='h'))
{
n[l]='e';
n[l+1]='s';
n[l+2]='\0';
}
else
{
n[l]='s';
n[l+1]='\0';
}
printf("%s",n);
return(0);
}
Input
INPUT : n - given string (word)
Output
OUTPUT:
printf("%s",n);
Processing
if(n[l-1]=='y')
{
n[l-1]='i';
n[l]='e';
n[l+1]='s';
n[l+2]='\0';
}
else if(n[l-1]=='s'||(n[l-2]=='c'&&n[l-1]=='h')||(n[l-2]=='s'&&n[l-1]=='h'))
{
n[l]='e';
n[l+1]='s';
n[l+2]='\0';
}
else
{
n[l]='s';
n[l+1]='\0';
}
Pseudocode
BEGIN
Read n
if(n[l-1]=='y')
{
n[l-1]='i';
n[l]='e';
n[l+1]='s';
n[l+2]='\0';
}
else if(n[l-1]=='s'||(n[l-2]=='c'&&n[l-1]=='h')||(n[l-2]=='s'&&n[l-1]=='h'))
{
n[l]='e';
n[l+1]='s';
n[l+2]='\0';
}
else
{
n[l]='s';
n[l+1]='\0';
}
printf("%s",n);
END
Question 4 (Count two digit prime numbers)
Given a number ‘n’, write an algorithm and the subsequent ‘C’ program to count the number of two digit prime numbers in it when adjacent digits are taken. For example, if the value of ‘n’ is 114 then the two digit numbers that can be formed by taking adjacent digits are 11 and 14. 11 is prime but 14 is not. Therefore print 1.
Input Format
A number
Output Format
Count of two digit prime numbers that can be formed when adjacent digits are taken
Solution
#include < stdio.h >
#include < string.h >
int prime(int n)
{
int i;
int flag=1;
for(i=n-1;i > 1;i--)
if(n%i==0)
{
flag=0;
break;
}
if(flag==1)
return(1);
else
return(0);
}
int main()
{
int i,n;
int c=0;
char s[20];
scanf("%s",s);
for(i=0;i < strlen(s)-1;i++)
{
n=(10*(s[i]-'0'))+(s[i+1]-'0');
c+=prime(n);
}
printf("%d",c);
return(0);
}
Input
INPUT: s - given number
Output
OUTPUT:
printf("%d",c);
Processing
int prime(int n)
{
int i;
int flag=1;
for(i=n-1;i > 1;i--)
if(n%i==0)
{
flag=0;
break;
}
if(flag==1)
return(1);
else
return(0);
}
Pseudocode
BEGIN
Read s
int prime(int n)
{
int i;
int flag=1;
for(i=n-1;i > 1;i--)
if(n%i==0)
{
flag=0;
break;
}
if(flag==1)
return(1);
else
return(0);
}
printf("%d",c);
END
Question 5 (Verification of ‘L’ shaped arrangement of coins on game board)
Consider a nxn board game with four types of coins red, green, blue and yellow. Given the state of the board with coins in all cells, develop an algorithm and write a C program to check if the same coins are placed in ‘L’ shape on the board. The number of cells in the vertical and horizontal line of ‘L’ shape should be same and should be greater than 1. Red coins are represented by ‘r’, blue coins are represented by ‘b’, green coins are represented by ‘g’ and yellow coins are represented by ‘y’.
For example, given the configuration of a 4 X 4 board with coins as shown below,
b r y r
r r y b
y r b b
b r r r
the program must print ‘Yes’. As there is a ‘L’ shape as shown in the figure below.
Input Format
Number of rows, r
Number of columns, c
Next ‘r x c’ lines contains the elements in the matrix
Elements in the matrix given row by row
Output Format
Print Yes or No
Solution
#include < stdio.h >
int main()
{
int r,n,i,j;
int f=0;
char c[10][10];
scanf("%d%d",&r,&n);
for(i=0;i < r;i++)
for(j=0;j < n;j++)
scanf("%c",&c[i][j]);
if(r > 2 && n > 2)
{
for(i=0;i < r-1;i++)
for(j=0;j < n-2;j++)
{
if((c[i][j]==c[i+1][j])&&(c[i][j]==c[i+1][j+1])&&(c[i][j]==c[i+1][j+2]))
{
f=1;
printf("Yes");
break;
}
}
if(f==0)
printf("No");
}
else
printf("No");
return(0);
}
Input
INPUT: r - number of rows n - number of columns c - matrix
Output
OUTPUT:
if(n > 2 &&r > 2)
{
if(f==0)
printf("No");
else
printf("Yes");
}
else
printf("No");
Processing
for(i=0;i < r-1;i++)
for(j=0;j < n-2;j++)
{
if((c[i][j]==c[i+1][j])&&(c[i][j]==c[i+1][j+1])&&(c[i][j]==c[i+1][j+2]))
{
f=1;
printf("Yes");
break;
}
}
Pseudocode
BEGIN
Read r,n and c
if(r > 2 && n > 2)
{
for(i=0;i < r-1;i++)
for(j=0;j < n-2;j++)
{
if((c[i][j]==c[i+1][j])&&(c[i][j]==c[i+1][j+1])&&(c[i][j]==c[i+1][j+2]))
{
f=1;
printf("Yes");
break;
}
}
if(f==0)
printf("No");
}
else
printf("No");
END
{ skill smuggler } 
can you please check the excess number problem code in practice problem 8 as its showing error
LikeLike
Thanks
LikeLike
code for question 1: showing error
excess numbers question; not getting output
LikeLike
Done
LikeLike