Write a c program to accept 3 digits and print all possible combinations from these digits?

Answer 1

Everyone i am ajay Verma and i answer this question, Write this programe in note pad:-

using System;

class MyClass

{

public static void Main(string [] args)

{

int n,d,result=1,sum=0;

Console.WriteLine("Enter any Number");

n=Convert.ToInt32(Console.ReadLine());

for(;n>0;)

{

d=n%10;

Console.WriteLine(d);

n=n/10;

}

}

}

Answer 2

#include

#include

#include

//My name is Prince Fred in Uganda, East Africa

// please if you like this, e-mail me atmuteesafred@yahoo.com or,

// call +256-712-134-996 to make me your friend.

// I used borland c++ 4.5 to compile.

void pw(int,char[]);

char *one[]={" "," one"," two"," three"," four"," five"," six"," seven",

"eight"," Nine"," ten"," eleven"," twelve"," thirteen"," fourteen","fifteen"," sixteen"," seventeen"," eighteen"," nineteen"};

char *ten[]={" "," "," twenty"," thirty"," forty"," fifty"," sixty","seventy"," eighty"," ninety"};

void main()

{

int n;

clrscr();

printf("Enter any 3 digit no: ");

scanf("%d",&n);

if(n<=0)

printf("Enter numbers greater than 0");

else

{

pw((n/100)," hundred and");

pw((n%100)," ");

}

getch();

}

void pw(int n,char ch[])

{

(n>19)?printf("%s %s ",ten[n/10],one[n%10]):printf("%s ",one[n]);

if(n)

printf("%s ",ch);

}

Answer 3

#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] z)

break;

for (int j = 0; j < s; j++)

cout<<a[j];

cout<<"n";

}

getch();

}

Answer 4

When dealing with a combination (in the mathematical sense), the order of the elements does not matter. Thus {1, 2, 3} is the exact same combination as {3, 2, 1}. The latter is known as a permutation (of the combination). For any given 3-digit combination there are 6 permutations, however if any two of the three digits are the same digit, then there are only 3 permutations, and if all three digits are the same then there can be only one permutation.

The following function will print all possible permutations of any 3 digits.

void print_permutations (int a, int b, int c) {

// sort digits in ascending order

if (b<a) a^=b^=a^=b; if (c<b) c^=b^=c^=b; if (b<a) a^=b^=a^=b;

if (a==b && b==c) { // all 3 digits match

printf ("{%d, %d, %d}\n", a, b, c);

} else if (a==b) { // first two digits match

printf ("{%d, %d, %d}\n", a, b, c);

printf ("{%d, %d, %d}\n", a, c, b);

printf ("{%d, %d, %d}\n", c, a, b);

} else if (b==c) { // last two digits match

printf ("{%d, %d, %d}\n", a, b, c);

printf ("{%d, %d, %d}\n", b, a, c);

printf ("{%d, %d, %d}\n", b, c, a);

} else { // all digits unique

printf ("{%d, %d, %d}\n", a, b, c);

printf ("{%d, %d, %d}\n", a, c, b);

printf ("{%d, %d, %d}\n", b, a, c);

printf ("{%d, %d, %d}\n", b, c, a);

printf ("{%d, %d, %d}\n", c, a, b);

printf ("{%d, %d, %d}\n", c, b, a);

}

}

To test the function, use the following:

int main () {

int a, b, c;

printf ("Enter three digits: ");

scanf ("%d %d %d", &a, &b, &c);

print_permutations (a, b, c);

return 0;

}