C/C++ Coding challenges

[Enigma_1]

you guys can use any method. @yrollgayanth well done machan. . elama thama

 

[Dileepa.]

you guys can use any method. @yrollgayanth well done machan. . elama thama

aiyoooo...

i thought we should use our own algo

 

[yrollgayanth]

aiyoooo...

i thought we should use our own algo

sure u can machan. if u want to find some trouble.

 

[yrollgayanth]

you guys can use any method. @yrollgayanth well done machan. . elama thama

Thanx machan

 

[Enigma_1]

aiyoooo...

i thought we should use our own algo

no machan. use any standard library function.

 

[madurax86]

well if there's one thing I can't stand thats algos lol here goes,

#include<stdio.h>
#include<stdlib.h>
inline void addhex(char *s, int num){
    switch (num){
        case 15: { *s = 'F'; break; }
        case 14: { *s = 'E'; break; }
        case 13: { *s = 'D'; break; }
        case 12: { *s = 'C'; break; }
        case 11: { *s = 'B'; break; }
        case 10: { *s = 'A'; break; }
        default: { *s = (char)48+num; break; }
    }
}
int main()
{
    unsigned long long int num, pnum, pos=0;
    char out[256]="";
    char out2[256]="";
    
    scanf("%lld", &num);
    
    while (num>16){ 
        pnum=num;
        num=num>>4;
        addhex( &out[pos], pnum-(num<<4));
        pos++;
    }
    addhex( &out[pos], num);
    for (num=0;num<=pos;num++) out2[num] = out[pos - num];
    printf("0x%s ", out2);
    return 0;
}

for those who are interested, heres the timings for decimal 999999999999 on a 2.0 GHz

999999999999 
0xE8D4A50FFF 
real	0m0.001s
user	0m0.000s
sys	0m0.004s

anyone can edit the code and make improvements,

 

[Enigma_1]

well if there's one thing I can't stand thats algos lol here goes,

#include<stdio.h>
#include<stdlib.h>
inline void addhex(char *s, int num){
    switch (num){
        case 15: { *s = 'F'; break; }
        case 14: { *s = 'E'; break; }
        case 13: { *s = 'D'; break; }
        case 12: { *s = 'C'; break; }
        case 11: { *s = 'B'; break; }
        case 10: { *s = 'A'; break; }
        default: { *s = (char)48+num; break; }
    }
}
int main()
{
    unsigned long long int num, pnum, pos=0;
    char out[256]="";
    char out2[256]="";
    
    scanf("%lld", &num);
    
    while (num>16){ 
        pnum=num;
        num=num>>4;
        addhex( &out[pos], pnum-(num<<4));
        pos++;
    }
    addhex( &out[pos], num);
    for (num=0;num<=pos;num++) out2[num] = out[pos - num];
    printf("0x%s ", out2);
    return 0;
}

for those who are interested, heres the timings for decimal 999999999999 on a 2.0 GHz

999999999999 
0xE8D4A50FFF 
real    0m0.001s
user    0m0.000s
sys    0m0.004s

anyone can edit the code and make improvements,

wow this is grt. fantastic machan. ela kiri.

 

[Dileepa.]

wow this is grt. fantastic machan. ela kiri.

madura is a great programmer machan!
most of the elakirians use his s/w for write sinhala.

http://www.madurax86.com/rsinglish/

 

[rclakmal]

public class HexConverter {

    private String s = "";
    private long number;

    public HexConverter(long number) {
        this.number = number;
    }

    public String convertToHex() {
        long remainder = number;
        while (remainder >= 16) {
            long x;
            x = remainder % 16;
            increment(x);
            remainder = remainder / 16;
        }
        increment(remainder);
        return s;
    }

    public void increment(long x) {
        if (x <= 9) {
            s = x+s;
        } else {
            char c = (char) (x + 55);
            s =c+s;
        }
    }

    public static void main(String[] args) {
        long start = System.nanoTime();
        System.out.println(new HexConverter(999999999999l).convertToHex());
        long end = System.nanoTime();
        System.out.println("Time Taken:" + (end - start) + "nanosecond");
    }
}

E8D4A50FFF
Time Taken:613558nanosecond

 

[rclakmal]

Anyway java has a inbuilt function to do this

long start = System.nanoTime();
 System.out.println(Long.toHexString(999999999999l);
  long end = System.nanoTime();
       System.out.println("Time Taken:"+(end-start)+"nanoseconds");

It takes

e8d4a50fff
Time Taken:524443nanoseconds

 

[Enigma_1]

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

int number;
char* HexNumber =NULL;

int main()
{

    printf("Please enter an integer :");
    scanf("%d",&number);
    HexNumber =(char*)malloc(50);
    sprintf(HexNumber,"%x",number);
    printf("1st method : hex representation is 0x%s\n",HexNumber);
    printf("2nd method : hex representation is 0x%p\n",number);
    printf("3rd method : hex representation is 0x%x\n",number);
    system("pause");
    return 0;
}

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

unsigned int number,modval,Divval,index=0;
char* HexNumber =NULL;
char* tem;

int main()
{
    printf("Please enter an integer :");
    scanf("%d",&number);
    HexNumber =(char*)malloc(50);
    tem =(char*)malloc(3);
    modval = number%16;
    Divval = (int)number/16;
    if(Divval<16)
    {
        sprintf(tem,"%d%d",Divval,modval);
        strcpy(HexNumber,tem);
        printf("Hex value is :%s\n",HexNumber);
        system("pause");
        exit(0);
    }
    else
    {
        sprintf(tem,"%d",modval);
        strcpy(HexNumber,tem);
        HexNumber = HexNumber+1;
        index++;
    }
    while(Divval>16)
    {
        modval = Divval%16;
        switch(modval)
        {
        case 10:
            strcpy(HexNumber,"A");
            break;
        case 11:
            strcpy(HexNumber,"B");
            break;
        case 12:
            strcpy(HexNumber,"C");
            break;
        case 13:
            strcpy(HexNumber,"D");
            break;
        case 14:
            strcpy(HexNumber,"E");
            break;
        case 15:
            strcpy(HexNumber,"F");
            break;
        default:
            sprintf(tem,"%d",modval);
            strcpy(HexNumber,tem);
            break;
        }
        HexNumber = HexNumber+1;
        Divval = (int)Divval/16;
        index++;
    }
    sprintf(tem,"%d",Divval);
    strcpy(HexNumber,tem);
    printf("The Hex value is 0x");
    while(strlen(HexNumber)!=(index+2))
    {
        printf("%c",HexNumber[0]);
        HexNumber--;
    }
    printf("\n");
    system("pause");
    return 0;
}

here are my solutions.

 

[Enigma_1]

madura is a great programmer machan!
most of the elakirians use his s/w for write sinhala.

http://www.madurax86.com/rsinglish/

yah 100% agree with you machan. he is a grt coder.

 

[DURApix]

hm...seems to be a nice thread..
elazzz...

 

[Enigma_1]

public class HexConverter {

    private String s = "";
    private long number;

    public HexConverter(long number) {
        this.number = number;
    }

    public String convertToHex() {
        long remainder = number;
        while (remainder >= 16) {
            long x;
            x = remainder % 16;
            increment(x);
            remainder = remainder / 16;
        }
        increment(remainder);
        return s;
    }

    public void increment(long x) {
        if (x <= 9) {
            s = x+s;
        } else {
            char c = (char) (x + 55);
            s =c+s;
        }
    }

    public static void main(String[] args) {
        long start = System.nanoTime();
        System.out.println(new HexConverter(999999999999l).convertToHex());
        long end = System.nanoTime();
        System.out.println("Time Taken:" + (end - start) + "nanosecond");
    }
}

E8D4A50FFF
Time Taken:613558nanosecond

Ah ela kiri machan.

 

[madurax86]

Anyway java has a inbuilt function to do this

long start = System.nanoTime();
 System.out.println(Long.toHexString(999999999999l);
  long end = System.nanoTime();
       System.out.println("Time Taken:"+(end-start)+"nanoseconds");

It takes

e8d4a50fff
Time Taken:524443nanoseconds

hey, nice seeing you again
BTW that time i gave was the process time,(=time it takes to load the program up + time it takes to run until it ends).

 

[madurax86]

yah 100% agree with you machan. he is a grt coder.

thanks

 

[DURApix]

Need a new challenge in C++...???

 

[rclakmal]

hey, nice seeing you again
BTW that time i gave was the process time,(=time it takes to load the program up + time it takes to run until it ends).

Same here machan ahhhh Yr I get it !! Anyway if you consider the same algorithm implemented in c++ and java ,java uses more resources because it runs on the JVM . Hoping to see more and more works from u in Elakiri !!!

 

[Enigma_1]

well if there's one thing I can't stand thats algos lol here goes,

#include<stdio.h>
#include<stdlib.h>
inline void addhex(char *s, int num){
    switch (num){
        case 15: { *s = 'F'; break; }
        case 14: { *s = 'E'; break; }
        case 13: { *s = 'D'; break; }
        case 12: { *s = 'C'; break; }
        case 11: { *s = 'B'; break; }
        case 10: { *s = 'A'; break; }
        default: { *s = (char)48+num; break; }
    }
}
int main()
{
    unsigned long long int num, pnum, pos=0;
    char out[256]="";
    char out2[256]="";
    
    scanf("%lld", &num);
    
    while (num>16){ 
        pnum=num;
        num=num>>4;
        addhex( &out[pos], pnum-(num<<4));
        pos++;
    }
    addhex( &out[pos], num);
    for (num=0;num<=pos;num++) out2[num] = out[pos - num];
    printf("0x%s ", out2);
    return 0;
}

for those who are interested, heres the timings for decimal 999999999999 on a 2.0 GHz

999999999999 
0xE8D4A50FFF 
real    0m0.001s
user    0m0.000s
sys    0m0.004s

anyone can edit the code and make improvements,

This is grt. I ran his code with a windows high resolution timer. results are outstanding. by considering all the other codes (including mine) I believe this is the fastest solution for the above challenge. . elama thama machan. keep it up.

 

[DURApix]

hello friends this is my answer to dat problem..
i used an integer array for the conversion process.


//(c)dURA
#include<iostream>
using namespace std;

int main()
{
int now,x,j;
begin:

const int size=20;
int no[size]={0};

cout<<"Please enter the integer value : ";
cin>>x;

now=x;
int i=0;
for(i=0;;i++)
{
if (now==0) break;
no=now%16;
now=now/16;
if (now<16 and now>0)
{i=i+1;no=now;break;}
}

cout<<"\nhexadecimal representation of "<<x<<" is 0x";

for(j=i;j>=0;j--)
{
if (no[j]>9)
{
switch (no[j]){
case 10 : cout<<"A";break;
case 11 : cout<<"B";break;
case 12 : cout<<"C";break;
case 13 : cout<<"D";break;
case 14 : cout<<"E";break;
case 15 : cout<<"F";break;
default : cout<<"Z";break;
}
}
else cout<<no[j];
}

cout<<"\n\t\t____\n\n";
goto begin;
}