Sum_of_Round_Numbers.cpp

/*Sum of Round Numbers
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A positive (strictly greater than zero) integer is called round if it is of the form d00...0. In other words, a positive integer is round if all its digits except the leftmost (most significant) are equal to zero. In particular, all numbers from 1 to 9 (inclusive) are round.

For example, the following numbers are round: 4000, 1, 9, 800, 90. The following numbers are not round: 110, 707, 222, 1001.

You are given a positive integer n (1≤n≤104). Represent the number n as a sum of round numbers using the minimum number of summands (addends). In other words, you need to represent the given number n as a sum of the least number of terms, each of which is a round number.

Input
The first line contains an integer t (1≤t≤104) — the number of test cases in the input. Then t test cases follow.

Each test case is a line containing an integer n (1≤n≤104).

Output
Print t answers to the test cases. Each answer must begin with an integer k — the minimum number of summands. Next, k terms must follow, each of which is a round number, and their sum is n. The terms can be printed in any order. If there are several answers, print any of them.

Example
inputCopy
5
5009
7
9876
10000
10
outputCopy
2
5000 9
1
7 
4
800 70 6 9000 
1
10000 
1
10 
*/
#include <bits/stdc++.h>
using namespace std;

int main(){
	int n;
	cin>>n;
  string st1;
  for(int i=0;i<n;i++){
    int sum=0;
    cin>>st1;
    for(int k=0;k<st1.size();k++){
      if(st1[k]>48){
        sum=sum+1;
      }
    }
    cout<<sum<<endl;

    int int1 = stoi(st1);
    for(int l=0;l<st1.size()-1;l++){
      int r=int1%10;
      if(r>0){
        cout<<(r*(pow(10,l)))<<" ";
      }
      int1=int1/10;
    }
    for(int m=st1.size()-1;m<st1.size();m++){
      int r=int1%10;
      if(r>0){
      cout<<(r*(pow(10,m)))<<endl;
      }
    }
  }
	return 0;
}