Write a C program to convert a Roman numeral to its decimal equivalent.

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<stdlib.h>

void main()
{

int *a,len,i,j,k;
char *rom;

clrscr();

printf(“Enter the Roman Numeral:”);
scanf(“%s”,rom);

len=strlen(rom);

for(i=0;i<len;i++)
{
if(rom[i]==’I')
a[i]=1;
else if(rom[i]==’V')
a[i]=5;
else if(rom[i]==’X')
a[i]=10;
else if(rom[i]==’L')
a[i]=50;
else if(rom[i]==’C')
a[i]=100;
else if(rom[i]==’D')
a[i]=500;
else if(rom[i]==’M')
a[i]=1000;
else
{
printf(“\nInvalid Value”);
getch();
exit(0);
}
}
k=a[len-1];
for(i=len-1;i>0;i–)
{
if(a[i]>a[i-1])
k=k-a[i-1];
else if(a[i]==a[i-1] || a[i]<a[i-1])
k=k+a[i-1];
}
printf(“\nIts Decimal Equivalent is:”);
printf(“%d”,k);
getch();
}

Write a C program to find the 2’s complement of a binary number.

2’s complement of a number is obtained by scanning it from right to left and complementing all the bits after the first appearance of a 1. Thus 2’s complement of 11100 is 00100. Write a C program to find the 2’s complement of a binary number.   Program: clrscr(); printf("Enter the binary number"); gets(a); for(i=0;a[i]!='\0'; i++) { if (a[i]!='0' && a[i]!='1') { printf("The number entered is not a binary number. Enter the correct number"); exit(0); } } complement(a); getch(); } void complement (char *a) { int l, i, c=0; char b[16]; l=strlen(a); for (i=l-1; i>=0; i--) { if (a[i]=='0') b[i]='1'; else b[i]='0'; } for(i=l-1; i>=0; i--) { if(i==l-1) { if (b[i]=='0') b[i]='1'; else { b[i]='0'; c=1; } } else { if(c==1 && b[i]=='0') { b[i]='1'; c=0; } else if (c==1 && b[i]=='1') { b[i]='0'; c=1; } } } b[l]='\0'; printf("The 2's complement is %s", b); } /* Output: Enter the binary number101 The 2's complement is 011 */

Write a C program to read in two numbers, x and n, and then compute the sum of this geometric progression: 1+x+x2+x3+………….+xn

Write a C program to read in two numbers, x and n, and then compute the sum of this geometric progression: 1+x+x2+x3+………….+xn For example: if n is 3 and x is 5, then the program computes 1+5+25+125. Print x, n, the sum Perform error checking. For example, the formula does not make sense for negative exponents – if n is less than 0. Have your program print an error message if n<0, then go back and read in the next pair of numbers of without computing the sum. Are any values of x also illegal ? If so, test for them too.   Program: #include<stdio.h> #include<conio.h> #include<math.h> void main() { int s_sum,i,x,n; clrscr(); printf("Enter the values for x and n:"); scanf("%d %d",&x,&n); if(n<=0 || x<=0) { printf("Value is not valid\n"); } else { s_sum=1; for(i=1;i<=n;i++) { s_sum=s_sum+pow(x,i); } printf("Sum of series=%d\n",s_sum); } getch(); } /* Output: Enter the values for x and n:2 4 Sum of series=31 */

Write a C program to construct a pyramid of numbers.

#include<stdio.h> #include<conio.h> void main() { int num,i,y,x=35; clrscr(); printf("\nEnter the number to generate the pyramid:\n"); scanf("%d",&num); for(y=0;y<=num;y++) { /*(x-coordinate,y-coordinate)*/ gotoxy(x,y+1); /*for displaying digits towards the left and right of zero*/ for(i=0-y;i<=y;i++) printf("%3d",abs(i)); x=x-3; } getch(); } /* Output: Enter the number to generate the pyramid: 5 0 1 0 1 2 1 0 1 2 3 2 1 0 1 2 3 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5

Write a C program to generate Pascal’s triangle.

#include<stdio.h> #include<conio.h> void main() { int bin,p,q,r,x; clrscr(); bin=1; q=0; printf("Rows you want to input:"); scanf("%d",&r); printf("\nPascal's Triangle:\n"); while(q<r) { for(p=40-3*q;p>0;--p) printf(" "); for(x=0;x<=q;++x) { if((x==0)||(q==0)) bin=1; else bin=(bin*(q-x+1))/x; printf("%6d",bin); } printf("\n"); ++q; } getch(); } /* Rows you want to input:6 Pascal's Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

Write a C program to count the lines, words and characters in a given text.

#include <stdio.h>
main()
{
char line[81], ctr;
int i,c,
end = 0,
characters = 0,
words = 0,
lines = 0;
printf(“KEY IN THE TEXT.\n”);
printf(“GIVE ONE SPACE AFTER EACH WORD.\n”);
printf(“WHEN COMPLETED, PRESS ‘RETURN’.\n\n”);
while( end == 0)
{
/* Reading a line of text */
c = 0;
while((ctr=getchar()) != ‘\n’)
line[c++] = ctr;
line[c] = ”;
/* counting the words in a line */
if(line[0] == ”)
break ;
else
{
words++;
for(i=0; line[i] != ”;i++)
if(line[i] == ‘ ‘ || line[i] == ‘\t’)
words++;
}
/* counting lines and characters */
lines = lines +1;
characters = characters + strlen(line);
}
printf (“\n”);
printf(“Number of lines = %d\n”, lines);
printf(“Number of words = %d\n”, words);
printf(“Number of characters = %d\n”, characters);
}

Output

KEY IN THE TEXT.
GIVE ONE SPACE AFTER EACH WORD.
WHEN COMPLETED, PRESS ‘RETURN’.
Admiration is a very short-lived passion.
Admiration involves a glorious obliquity of vision.
Always we like those who admire us but we do not
like those whom we admire.
Fools admire, but men of sense approve.
Number of lines = 5
Number of words = 36
Number of characters = 205

a) Write a C program that displays the position or index in the string S where the string T begins, or – 1 if S doesn’t contain T.

#include<stdio.h> #include<string.h> #include<conio.h> void main() { char s[30], t[20]; char *found; clrscr(); /* Entering the main string */ puts("Enter the first string: "); gets(s); /* Entering the string whose position or index to be displayed */ puts("Enter the string to be searched: "); gets(t); /*Searching string t in string s */ found=strstr(s,t); if(found) printf("Second String is found in the First String at %d position.\n",found-s); else printf("-1"); getch(); } /* Output: Enter the first string: sridhar siricilla Enter the string to be searched: siri Second String is found in the First String at 8 position. */