Thread: Integer analysis, question

Design and code a program that will take in an integer number from the user. This number has to be between 1 and 9999 (inclusive). This number will then be analysed by your code as to how many 1000s, 100s, 10s, and 1s make up this number. For example, an entry of 4208 by the user should make your program output something like: 4208 is made up of 4 lots of 1000, 2 lots of 100, 0 lots of 10s, and 8 lots of 1s.
So 4208 = 4 x 1000 + 2 x 100 + 0 x 10 + 8 x 1

Another example is 233. The output should be something like “0 lots of 1000, 2 lots of 100, 3 lots of 10s, and 3 lots of 1s”.

Can i please have help with writing this code, and any examples please.

This is how we have started:

Code:

#include <stdio.h>

main()
{
      int user_in;
      
      
      printf("enter an integer from value 1 to 9999:\n\n");
      scanf("%d", &user_in);
      
      
      if (user_in < 1000 > 0);
        printf("0 x 1000\n"); 
      if (user_in < 2000 > 999);
        printf("1 x 1000\n");
      if (user_in < 3000 > 1999);
        printf("2 x 1000\n");
      if (user_in < 4000 > 2999);
        printf("3 x 1000\n");
      if (user_in < 5000 > 3999);
        printf("4 x 1000\n");
      if (user_in < 6000 > 4999);
        printf("5 x 1000\n");
      
      
      
      
      
      
      
      system("pause");
      }

thanks

Thanks buddy,

Can you show me the fastest and simplest way you would do it please.

Probably the best way is to sit down with a pencil and paper and work out how you would dissect a number... then look in the C documentation to see if there are math functions that will help you achieve that goal...

Quzah already mentioned modulous math (%) ... what is the result of 103 mod 10 ... 3, right? That should give you a pretty strong hint.

Two suggestions for you.

1) If you want to use "stair step" type logic, it's usually far better to start with the greatest value, and step down in size, from there, to the smallest value.

2) If you have a number, say 123, you can easily "peel" off the digits one at a time, by using this:

Code:

number=123
count = 0
while(number > 0) 
   count++;
   digit = number mod 10
   number /= 10;
end of while loop

When loop is 1, your digit will be 1's digit. When it's 2, your digit is the 10's digit, etc.

Makes things a lot easier, imo.

Code:

      if (user_in < 6000 > 4999);

This!
a < b > c is not C. you've to use && operator a < b && b > c.
and if( A ) ; is not right.
It's the same as

Code:

if(A) {  // do nothing  }

I am also attacking this question, I see that the modulus (%) can be used but could there please be a post to start the code of as i have tried for hours and have had no success! thanks

We are still having a few issues, do you think this is the best way to do this problem?

Thanks guys

Code:

#include <stdio.h>

main()
{
int user_in;


printf("************* enter an integer from value 1 to 9999: ******************\n\n");
scanf("%d", &user_in);


if(user_in<1000){
printf("0 x 1000\n"); 
} 
else
if((user_in < 2000)&&(user_in >999)){
printf("1 x 1000\n"); 
} 
else
if((user_in < 3000)&&(user_in >1999)){
printf("2 x 1000\n"); 
}
else
if((user_in < 4000)&&(user_in >2999)){
printf("3 x 1000\n"); 
}
else
if((user_in < 5000)&&(user_in >3999)){
printf("4 x 1000\n"); 
}
else
if((user_in < 6000)&&(user_in >4999)){
printf("5 x 1000\n"); 
} 
else
if((user_in < 7000)&&(user_in >5999)){
printf("6 x 1000\n"); 
} 
else
if((user_in < 8000)&&(user_in >6999)){
printf("7 x 1000\n"); 
}
else
if((user_in < 9000)&&(user_in >7999)){
printf("8 x 1000\n"); 
}
else
if((user_in < 10000)&&(user_in >8999)){
printf("9 x 1000\n"); 
}


else
printf("**************** re-enter number *****************");

system("pause");
}

Try it like this...

Code:

#include <stdio.h>

int main (void)
  { int digit;
     int multiplier = 1;      
     unsigned int number;     // lets us go up to 9 digits


     // get number from the operator

    while (number > 0)
      { digit = number % 10
         printf("%d x %d\n",digit, multiplier);
         number /= 10;
         multiplier *= 10; }

   return 0; }

hi, thanks for the help this is so far what we have written but there must be a way to short cut this otherwise we would be writting a 1500 line code by the time we do this for the 1s units.

Code:

#include <stdio.h>

main()
{
      int user_in;
      
      
      printf("*************  enter an integer from value 1 to 9999: ******************\n\n");
      scanf("%d", &user_in);
      
      
      if(user_in<1000){
       printf("0 x 1000\n"); 
      } 
      else
       if((user_in < 2000)&&(user_in >999)){
      printf("1 x 1000\n"); 
      }            
      else
       if((user_in < 3000)&&(user_in >1999)){
      printf("2 x 1000\n"); 
      }
      else
       if((user_in < 4000)&&(user_in >2999)){
      printf("3 x 1000\n"); 
      }
      else
       if((user_in < 5000)&&(user_in >3999)){
      printf("4 x 1000\n"); 
      }
      else
       if((user_in < 6000)&&(user_in >4999)){
      printf("5 x 1000\n"); 
      }      
      else
       if((user_in < 7000)&&(user_in >5999)){
      printf("6 x 1000\n"); 
      }      
      else
       if((user_in < 8000)&&(user_in >6999)){
      printf("7 x 1000\n"); 
      }
      else
       if((user_in < 9000)&&(user_in >7999)){
      printf("8 x 1000\n"); 
      }
      else
       if((user_in < 10000)&&(user_in >8999)){
      printf("9 x 1000\n"); 
      }
      
      
      else{
          printf("****************  re-enter number  *****************");
          }
      system("pause");
      }

hi,
thanks for the help we have now after some time figured out a suitable way to answer this, only that we now want the program to run continuously until stopped by the user, any suggestions would be aprreciated.

Code:

 #include <stdio.h>

main()
{
      int user_in;
      
      printf("*************  enter an integer from value 1 to 9999: ******************\n\n");
      scanf("%d", &user_in);
      
      int a  = user_in/1000;
      int b = user_in % 1000 / 100;   
      int c = user_in % 100 / 10;    
      int d = user_in % 10 / 1;    
      
      
      printf("%d lots of 1000s\n\n",a);
      
      printf("%d lots of 100s\n\n",b);
      
      printf("%d lots of 10s\n\n",c);
      
      printf("%d lots of 1s\n\n",d); 
 
 
     
 
      
      if (user_in > 9999) {
                 printf("error");
                 }
      
      
      system("pause");
      }