Yanesh Tyagi

• Dec 23rd, 2006

 

/* Program to write combination of 3 digit number
   Author : Rachit Tyagi
   Date : 23-Dec-2006 2345 IST
*/
 
#include<iostream.h>
#include<conio.h>
void main(){
 clrscr();
 const int s = 3;
 int a[s];
 int x,y,z;
 for (int i=0; i<s; i++)
  cin >> a[i];
  x = a[0];
  y = a[1];
  z = a[2];
  i = 0;
  cout<<x<<y<<z;
  cout<<"n";
  while(1){
  a[s] = a[i];
  if (i != s - 1){
   a[i] = a[i+1];
   a[i+1] = a[s];
  }
  else {
   a[i] = a[0];
   a[0] = a[s];
   i=-1;
  }
  i++;

  if (a[0] == x && a[1] == y && a[2] == z)
   break;

  for (int j = 0; j < s; j++)
   cout<<a[j];
  cout<<"n";

  }
  getch();

}

manochasunil

 Profile Answers by manochasunil 

• Feb 12th, 2007

 

int a[3]; a[0] = 1; a[1] = 2; a[2] = 3; // Print all the permutations of a three digit number for (int i = 0; i<=2;i++) { for (int j = 0; j<=2;j++) { if(i!=j) { printf("%d",a[i]); printf("%d",a[j]); for (int k = 0; k<=2;k++) { if((i!=k)&& j!=k) { printf("%dn",a[k]); } } } } }

sreerag

• Jul 3rd, 2013

 

Code
#include<iostream.h>


void main()


{


        int z,a,b,c,q,w,e,r,t;


        cout<<"enter a three digit number";


        cin>>z;


        a=z/100;


        c=z%10;


        b=(z-(a*100+c))/10;


        z=a*100+b*10+c;


        cout<<"first="<<z<<"


";


        q=a*100+c*10+b;


        cout<<"second="<<q<<"


";


        w=b*100+c*10+a;


        cout<<"third="<<w<<"


";


        e=b*100+a*10+c;


        cout<<"fourth="<<e<<"


";


        r=c*100+a*10+b;


        cout<<"fifth="<<r<<"


";


        t=c*100+b*10+a;


        cout<<"sixth="<<t;


}

adnan

• Jan 13th, 2016

 

Code
Yanesh Tyagi


 Dec 23rd, 2006


 


/* Program to write combination of 3 digit number


   Author : Rachit Tyagi


   Date : 23-Dec-2006 2345 IST


*/


 


#include<iostream.h>


#include<conio.h>


void main(){


 clrscr();


 const int s = 3;


 int a[s];


 int x,y,z;


 for (int i=0; i<s; i++)


  cin >> a[i];


  x = a[0];


  y = a[1];


  z = a[2];


  i = 0;


  cout<<x<<y<<z;


  cout<<"n";


  while(1){


  a[s] = a[i];


  if (i != s - 1){


   a[i] = a[i+1];


   a[i+1] = a[s];


  }


  else {


   a[i] = a[0];


   a[0] = a[s];


   i=-1;


  }


  i++;


  if (a[0] == x && a[1] == y && a[2] == z)


   break;


  for (int j = 0; j < s; j++)


   cout<<a[j];


  cout<<"n";


  }


  getch();


}

Lineesh K Mohan

• Mar 22nd, 2016

 

Code
#include<stdio.h>


#include<iostream>


 


using namespace std;


 


int main()


{


        int arrNum[3] = {0};


        int nTemp = 0;


        int nI = 0,nJ = 0,nK = 0;


 


        cout<<"


 Enter the 3 digit numbers      ";


        cin>>nTemp;


 


        arrNum[0] = nTemp/100;


        arrNum[1] = (nTemp/10) - (arrNum[0] * 10);


        arrNum[2] = nTemp%10;


 


        for(nI = 0; nI < 3; nI++)


        {


                nJ = (nI+1) % 3;


                nK = (nI+2) % 3;


                nTemp = 100 * arrNum[nI] + 10 * arrNum[nJ] + arrNum[nK];


                cout<<nTemp<<endl;


                nTemp = 100 * arrNum[nI] + 10 * arrNum[nK] + arrNum[nJ];


                cout<<nTemp<<endl;


        }


        return 0;


}