Qiuz Problems (Which Question you Got?)
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ShadowMan wrote: My Friends got
Problem : SetBits
You are given two characters. You Write a program to change the bits of 2 characters.
Method setbits has four parameters x,y,p and n, x and y are 2 characters and p is the position and n is the number of bits. Function should returns c which is a copy of x with the n bits from pth position of x replaced with the rightmost n bits of y leaving the other bits unchanged. If the value of p or n is greater than 8 bit then return as 0.
Example 1:
Input
unsigned char x = Q
unsigned char y = K
int p = 4
int n = 3
Output
New character c = M
Bit Representation of 'Q' = 01010001 and 'K' = 01001011, 3 bits from 4th position of 'Q' replaced with the rightmost 3 bits of 'K' leaving the other bits unchanged the bit representation is 01001101. The character for the new bit is 'M'. Hence 'M' is the output.
Example 2:
Input
unsigned char x = %
unsigned char y = +
int p = 5
int n = 2
Output
New character c = -
[code]
#include"Setbits.h"
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
void strpad(char str[], int len) {
int i;
strrev(str); for(i=len ; i<8 ; i++) str
='0';
str
='\0';
strrev(str);
}
unsigned char setbits(unsigned char x,int p,int n,unsigned char y) {
char xstr[10]={0}, ystr[10]={0};
char out[10]={0}, c=0;
int i,j;
if( p<=0 || p>8 || n<=0 || n>8) return 0;
itoa(x,xstr,2); itoa(y,ystr,2);
if(strlen(xstr) >8 || strlen(ystr)>8) return 0;
strpad(xstr, strlen(xstr)); strpad(ystr, strlen(ystr));
strcpy(out,xstr);
for(i=p-1, j=strlen(ystr)-n ; i<(p-1)+n ; i++, j++)
out = ystr[j];
if(strlen(out)>8) return 0;
for(i=7, j=0 ; i>=0 ; i--, j++) c= c + ((out-'0') * (int)(pow(2,j)));
return c;
}
void dsmain() {
unsigned char x='Q',y='K', c; int p=4,n=3;
printf("Output = [%c]\n",setbits(x,p,n,y));
}
[/code]
All replies
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ShadowMan wrote: My Friends got
Problem : SetBits
You are given two characters. You Write a program to change the bits of 2 characters.
Method setbits has four parameters x,y,p and n, x and y are 2 characters and p is the position and n is the number of bits. Function should returns c which is a copy of x with the n bits from pth position of x replaced with the rightmost n bits of y leaving the other bits unchanged. If the value of p or n is greater than 8 bit then return as 0.
Example 1:
Input
unsigned char x = Q
unsigned char y = K
int p = 4
int n = 3
Output
New character c = M
Bit Representation of 'Q' = 01010001 and 'K' = 01001011, 3 bits from 4th position of 'Q' replaced with the rightmost 3 bits of 'K' leaving the other bits unchanged the bit representation is 01001101. The character for the new bit is 'M'. Hence 'M' is the output.
Example 2:
Input
unsigned char x = %
unsigned char y = +
int p = 5
int n = 2
Output
New character c = -
[code]
#include"Setbits.h"
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
void strpad(char str[], int len) {
int i;
strrev(str); for(i=len ; i<8 ; i++) str='0';
str='\0';
strrev(str);
}
unsigned char setbits(unsigned char x,int p,int n,unsigned char y) {
char xstr[10]={0}, ystr[10]={0};
char out[10]={0}, c=0;
int i,j;
if( p<=0 || p>8 || n<=0 || n>8) return 0;
itoa(x,xstr,2); itoa(y,ystr,2);
if(strlen(xstr) >8 || strlen(ystr)>8) return 0;
strpad(xstr, strlen(xstr)); strpad(ystr, strlen(ystr));
strcpy(out,xstr);
for(i=p-1, j=strlen(ystr)-n ; i<(p-1)+n ; i++, j++)
out = ystr[j];
if(strlen(out)>8) return 0;
for(i=7, j=0 ; i>=0 ; i--, j++) c= c + ((out-'0') * (int)(pow(2,j)));
return c;
}
void dsmain() {
unsigned char x='Q',y='K', c; int p=4,n=3;
printf("Output = [%c]\n",setbits(x,p,n,y));
}
[/code] -
As you have not specified the details/prototypes of the functions to use, i have written the code entirely in main. may be you can identify it yoruself. and btw, this is the code for infix to postfix. i guess infix to prefix wouldn't be any trouble for you after understanding this.
[code]
#include<stdio.h>
#include<conio.h>
#include<ctype.h>
int PRIORITY(char ch)
{
if (ch=='(' || ch==')')
return 4;
else if (ch=='^')
return 3;
else if (ch=='*' || ch=='/')
return 2;
else if (ch=='+' || ch=='-')
return 1;
else
return 0;
}
void main()
{
char stack[40],p[40];
char str[40]={0};
int top=-1,p_cnt=0;
int i,j,k,len;
clrscr();
printf("Enter any Infix Equation : ");
gets(str);
printf("\nSymbol\t\tStack\t\tPostfix");
len=strlen(str);
str[len]=')';
str[len+1]='\0';
for (i=0 ; str ; i++)
{
if (i==0)
{
stack[++top]='(';
printf("\n\t\t(\n\r");
}
printf("%c\t\t",str);
if ( isalnum(str) )
{
p[p_cnt++]=str;
}
else
{
if (str!=')')
{
if (PRIORITY(str) > PRIORITY(stack[top]) || (stack[top]=='(') )
{
stack[++top]=str;
}
else
{
p[p_cnt++]=stack[top];
stack[top]=str;
}
}
else if (str==')')
{
while (stack[top]!='(')
{
p[p_cnt++]=stack[top--];
}
top--;
}
}
for (j=0 ; j<=top ; j++)
{
printf("%c",stack[j]);
}
printf("\t\t");
for (j=0 ; j<p_cnt ; j++)
{
printf("%c",p[j]);
}
printf("\n");
}
printf("\nPostFix Notation : ");
for (j=0 ; j<p_cnt ; j++)
{
printf("%c",p[j]);
}
getch();
}
[/code]
please mark as answer and give reps...... -
Adjective series is the product of digits giving the previous value of the series.For a given input value return the perfect and nearest adjective number series until it exists. Example 1: Input: unsigned int value = 6 Output: Returns {6, 16, 28, 47} Explanation: 6 can be product of 1 and 6 and 2 and 3. Hence the next possible number in the series could be 16 or 23 or 32 or 61. Since 16 is the smallest number of all the possible numbers it will be the next number in the series. Similarly 16 can be the product of 2 and 8 or 4 and 4. The possible numbers are 28, 82, 44. Since 28 is the smallest number it will be the next number in the series. By the same logic 47 is the next number. After 47 there is no number possible. Hence the series returned is {47, 28, 16, 6}. Example 2: Input: unsigned int value = 12 Output: Returns {12, 26}
Solution#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define INPUT 6
/* change INPUT value to test with different INPUTS */
unsigned getNum(unsigned int n1,unsigned int n2);
int* getAdjective(int v);
int *series; /* Holds series */
void dsmain()
{
unsigned int z;
z = INPUT; /* Change value of INPUT directive to different inputs */
series = getAdjective(z);
/* Display the series */
for(z=0;series![[Z]](/SamVaad//emoticons/emotion-47.gif)
!= -1;z++)
{
printf("%d\n",series);
}
}
/*Creates Single integer from two numbers passed */
unsigned getNum(unsigned int n1,unsigned int n2)
{
unsigned tmp,n2_digits=0;
tmp = n2;
while(tmp != 0)
{
n2_digits++;
tmp /= 10;
}
n1 = n1 * (unsigned int)pow(10,n2_digits)+n2;
return n1;
}
/* return AdjectiveSeries for given value*/
int* getAdjective(int v){
static int cnt = 0;
int z;
unsigned int x,a,smallest;
smallest = 65000;
if(cnt == 0)
{
series = (int*)malloc(sizeof(int)*20);
series[0] = v;
cnt++;
}
for(x=1;x*x < v;x++)
{
if(!(v%x))
{
if(getNum(x,v/x) <= getNum(v/x,x))
{
a = getNum(x,v/x);
}
else
{
a = getNum(v/x,x);
}
if(a < smallest)
{
smallest = a;
}
}
}
if(smallest < 99)
{
series[cnt++] = smallest;
getAdjective(smallest);
}
else
{
series[cnt++] = -1;
}
return series;
}
