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Problem description:
Known to be a positive integer n, n is the range of 1-999999999, your task is to put the integer factorization to a single number, and then from left to right, in turn, printing out each number, such as integer "12345", "1, 2, 3, 4, 5",

The input and output requirements:
Input consists of a positive integer, that is, to break up the integer n, 1 & lt;=n<=999999999, the output of integer resolution as a result, there is a space between the adjacent two Numbers, after the last digit is a newline character, such as separation results of the 12345 "1, 2, 3, 4, 5"

Program running effect:
Sample 1:
Input: 12345
Output: 1, 2, 3, 4, 5
Sample 2:
Input: 1
Output: 1

CodePudding user response:

 # include & lt; Stdio. H> 
# include & lt; String. H> 

Int main () 
{
Char num [10]. 
Memset (num, 1, sizeof (char) * 10); 
The scanf (" % s ", num); 
for (int i=0; i <10; I++) 
{
If (num [I]!=1) 
Printf (" % c ", num [I]); 
} 
printf("\n"); 
return 0; 
}

For your reference

CodePudding user response:

The

reference 1/f, 5250 response:

 # include & lt; Stdio. H> 
# include & lt; String. H> 

Int main () 
{
Char num [10]. 
Memset (num, 1, sizeof (char) * 10); 
The scanf (" % s ", num); 
for (int i=0; i <10; I++) 
{
If (num [I]!=1) 
Printf (" % c ", num [I]); 
} 
printf("\n"); 
return 0; 
}

For your reference

This program, you can do it:

 for (int I=0; i  CodePudding user response: 
 # include & lt; Stdio. H> 
# include & lt; Stdlib. H> 
# include & lt; String. H> 



Void print_digit (unsigned long num); 

Int main (void) 
{
Unsigned long num. 

The scanf (" % lu ", & amp; Num); 


Print_digit (num); 
Putchar (10); 

return 0; 
} 

Void print_digit (unsigned long num) 
{
If (num==0) 
return; 

Print_digit (num/10); 
Printf (" % lu ", num % 10); 
} 
 

For your reference ~ 

As above, can consider to use recursive functions CodePudding user response: 
  reference 2 building self-confidence boy reply:     Quote: reference 1/f, 5250 response:    # include & lt; Stdio. H> 
# include & lt; String. H> 

Int main () 
{
Char num [10]. 
Memset (num, 1, sizeof (char) * 10); 
The scanf (" % s ", num); 
for (int i=0; i <10; I++) 
{
If (num [I]!=1) 
Printf (" % c ", num [I]); 
} 
printf("\n"); 
return 0; 
} 
 

For your reference 
  

This program, you can do it: 
 for (int I=0; i  
  

thank you CodePudding user response: 
Thank you for your attention!