Write a c++ program that reads numbers from the user, and prints a table showing how many times each digit appears in each number. The user should be able to enter more than one number to be tested for repeated digits. The program should terminate when the user enters a number that is less than 0. Here is a sample output:
Enter a number (-1 to end this loop): 41271092
Digit: 0 1 2 3 4 5 6 7 8 9
Occurrences: 1 2 2 0 1 0 0 1 0 1
Enter a number (-1 to end this loop): 21272092
Digit: 0 1 2 3 4 5 6 7 8 9
Occurrences: 1 1 4 0 0 0 0 1 0 1
Enter a number (-1 to end this loop): 222345
Digit: 0 1 2 3 4 5 6 7 8 9
Occurrences: 0 0 3 1 1 1 0 0 0 0
Enter a number (-1 to end this loop): 98798
Digit: 0 1 2 3 4 5 6 7 8 9
Occurrences: 0 0 0 0 0 0 1 2 2
Enter a number (-1 to end this loop): -1
Here to store digit count array is used. Each index of array stores corresponding digit occurence .
C++ CODE
#include<iostream>
using namespace std;
// COUNT THE OCCURANCE DIGITS OF A NUMBER , USING ARRAY AND ITS
INDEX POSITION
//IF A DIGIT 0 IS FOUND THEN 0 INDEX OF ARRAY WILL INCRMENT. THEN
PRINT THE ARRAY
int main()
{
int num; // NU,BER
int x[10]; //ARRAY TO STORE DIGIT COUNT
int i,d; // TEMPORARY VAR
while(1) // LOOP UNTIL USER ENTERS -1
{
cout<<"\nEnter a numbber :
";
cin>>num;
if(num==-1) // CHECK IF -1 THEN
LOOP BREAKS
break;
for(i=0; i<10; i++) //
INITALLY ASSIGN 0 TO EACH ARRAY ELEMENTS
x[i]=0;
while (num> 0)
{
d=num % 10; //
GET INDIVIDUAL DIGIT OF NUMBER
x[d]++; //
INCREMENT SAME INDEX OF ARRAY, IF DIGIT GET 1 THEN 1 INDEX WILL BE
INCREMENTED REPEATEDLY
num = num / 10;
// DECREASE THE NUMBER FROM RIGHT SIDE
}
cout<<"\nDigits : ";
for(i=0; i<10; i++) // PRINT
DIGITS
cout<<i<<" ";
cout<<"\nOccurrences :
";
for(i=0; i<10; i++) // PRINT
ARRAY THAT CONTAINS THE DIGIT COUNT
cout<<x[i]<<" ";
}
return 0;
} // MAIN END
SCREEN SHOT CODE

SCREEN SHOT OUTPUT

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