/*
Write a program that accepts two strings.
* the program checks whether the second word is present anywhere in any word.
* it the word is found then it prints the word from second string.
*
* An example could be:
* Input :
* Entr first string: this is a iss thiss. al a/
* Enter second string: is
* output: this is iss thiss.
*/
/*
* File: main.c
* Author: umang
*
* Created on November 11, 2011, 3:08 PM
*/
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#define true 0
#define false 1
//int main(int argc, char** argv)
void main()
{
char arr[100] = {'\0'}; // initialize whole array
char word[10] = {'\0'};
int i, j, temp;
int b = false;
printf("Enter string: ");
gets(arr);
printf("Enter word: ");
gets(word);
for (i = 0; arr[i] != '\0'; i++)
{
temp = i;
b = false;
for (j = 0; word[j] != NULL && arr[j + i] != NULL; j++)
{
if (word[j] != arr[i + j])
{
b = true;
}
}
if (b == false)
{
j--;
if (arr[i + j] != ' ')
{
while (arr[i + j] != ' ' && arr[i + j] != NULL)
{
i++;
}
}
while (arr[temp] != ' ' && temp >= 0)
{
temp--;
if (temp < 0)
{
temp = 0;
break;
}
}
if (temp != 0)
{
temp++;
}
for (; arr[temp] != ' ' && arr[temp] != NULL; temp++)
{
printf("%c", arr[temp]);
}
printf(" ");
i = temp;
}
}
//return (EXIT_SUCCESS);
}
// input
// this a te is thisi is. t
// 4 12 18 22
Friday, November 11, 2011
string program in C
Saturday, October 1, 2011
no of coin problem
/*
rita has a money pouch containing Rsx (for example 700). There are equal number of 25 paise coins,
50 paise coins and one rupee coins. Write a C program to find out how many of each are there .
*/
/**************************
* author: umang bhatt *
* bhatt.umang7@gmail.com *
**************************/
/*
what i ubderstood:
0.25 x + 0.50 x + 1 x = amt
so dividing everything by 0.25 we get
x + 2x + 4x = amt * 4
therefore 7x = 4 amt
*/
#include<stdio.h>
void main()
{
int amt =700 ;
int i;
printf("Enter amount: " );
scanf("%d",&amt);
for( i = 1 ; i <= amt * 4 ; i++)
{
if( (7*i) == (4*amt) )
{
printf("no of coins are: %d\n",i );
break;
}
else if( (7*i) > (4*amt) )
{
printf("no such combination of coins can exist\n" );
break;
}
}
// printf("\n%f\n", ( (0.25*i) + (0.50 * i ) + ( i) ));
}
number operation
/************************** * author: umang bhatt * * bhatt.umang7@gmail.com * **************************/ /* consider following number 45 * 45 = 2025; 20+25 = 45. Write a program to generate number between 32 and 99 the satisfies the abovce property. */ #include<stdio.h> #define true 0 #define false 1 int foo(int n) { int i = n * n; int j ; int return_val; j = i % 100; i = i / 100; if((i+j)==n) { return_val = true; } else { return_val = false; } return return_val; } void main() { int i ; for( i = 32 ; i <= 99 ; i++) { if(foo(i) == true ) { printf("%d ",i ); } } printf("\n"); }
Sunday, September 11, 2011
number program
/*
Definition from Aniruddha Guddi:
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
*/
/**************************
* author: umang bhatt *
* bhatt.umang7@gmail.com *
**************************/
#include<stdio.h>
#define TRUE 0
#define FALSE 1
int contains_same_digits(int no1, int no2)
{
int arr1[10] = {0};
int arr1_top = -1;
int arr2[10] = {0};
int arr2_top = -1;
int temp = no1;
int i, j, k;
int returnval;
int flag = FALSE;
//store numbers in array
while (temp > 0)
{
arr1[++arr1_top] = temp % 10;
temp = temp / 10;
}
temp = no2;
while (temp > 0)
{
arr2[++arr2_top] = temp % 10;
temp = temp / 10;
}
for (i = 0; i <= arr1_top; i++)
{
flag = FALSE;
for (j = 0; j <= arr2_top; j++)
{
if (arr1[i] == arr2[j])
{
flag = TRUE;
// remove that element from both arrays
for (k = i; k < arr1_top; k++)
{
arr1[k] = arr1[k + 1];
}
arr1_top--;
for (k = j; k < arr2_top; k++)
{
arr2[k] = arr2[k + 1];
}
arr2_top--;
if (j != 0)
j--;
}
}
if (flag == TRUE)
{
i = -1;
}
}
if (arr1_top == -1 && arr2_top == -1)
{
returnval = TRUE;
}
else
{
returnval = FALSE;
}
return returnval;
}
int main()
{
char c[5];
int i, j;
int flag;
int flag1;
for (j = 2; j <= 6; j++)
{
i = 1;
flag = FALSE;
while (flag == FALSE)
{
flag = contains_same_digits(i, (i * j));
if (flag == TRUE)
{
printf("\n %d * %d = %d where digits of %d and %d are same ", i, j, (i * j), i, (i * j));
flag = TRUE;
}
i++;
}
}
printf("\n");
return 0;
}
C program : largest palindrome made by multiplication of two 3 digit numbers
/*
Definition from Aniruddha Guddi:
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
*/
/**************************
* author: umang bhatt *
* bhatt.umang7@gmail.com *
**************************/
#include<stdio.h>
#define TRUE 0
#define FALSE 1
int is_palindrome(int no)
{
int temp = no;
int ans = 0;
int r;
while (temp > 0)
{
ans = ((ans * 10) + (temp % 10));
temp = temp / 10;
}
if (ans == no)
{
r = TRUE;
}
else
{
r = FALSE;
}
return r;
}
int main()
{
int f_no = -1;
int l_no = -1;
int last_palindrom = -1;
int flag = FALSE;
int i, j;
for (i = 100; i < 999; i++)
{
for (j = 100; j < 999; j++)
{
if (is_palindrome(i * j) == TRUE)
{
if ((i * j) > last_palindrom)
{
last_palindrom = i * j;
f_no = i;
l_no = j;
}
}
}
}
printf("%d * %d = %d \n", f_no, l_no, last_palindrom);
return 0;
}
Tuesday, May 3, 2011
factorial using stack
/****************************
* Umang B Bhatt *
* bhatt.umang7@gmail.com *
*****************************/
/**
* program for factorial using stack
*/
#include<stdio.h>
#include<string.h>
#define MAX 100
// push method
int push(int *s, int top, int ele)
{
int i;
if (top >= MAX)
{
printf("\nStack Overflow");
}
else
{
s[++top] = ele;
}
return top;
}
//pop method
int pop(int *a, int *top)
{
if ((*top) >= 0)
{
(*top) = (*top) - 1;
return a[(*top) + 1];
}
else
{
printf("Stack underflow\n");
return 0;
}
}
// personally, i believe that declaring TOP and S here is bad practice
void main()
{
int n;
int i; // loop variable
int ans = 1 ; // stroes the final answer
int TOP = -1; // stack variables that mainntain stack's top
int s[MAX]; // the stack
printf("\nEnter number: ");
scanf("%d",&n);
// here we can also make sure that the use is not entering a number
// that can not be accomodated in the stack
// if the user enters a number less than or equal to 0
// then we can not find factorial
if(n<=0)
{
printf("\nThe number can not be less than 0");
}
else
{
// push the numbers n,n-1 ....1 in the stack
// you can also go in reverse manner here. the result will be the same
// because in multiplication, order does not matter
for(i = n ; i >0 ; i--)
{
TOP = push(s,TOP, i);
}
// now pop all the elements one by one
// multiply them with the answer variable
while(TOP>=0)
{
ans = ans * pop(s,&TOP);
}
printf("\nFactorail is %d\n",ans);
}
// getch(); // will not work on linux
}
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