题目描述

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.

样例

5
5009
7
9876
10000
10

2
5000 9
1
7 
4
800 70 6 9000 
1
10000 
1
10 

这道题很简单 模拟她的过程即可 注意string s.size() 和从0开始即可

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        string s;
        cin>>s;
        int num=0;
        for(int i=0;i<s.size();i++)
        {
            if(s[i]>'0'&&s[i]<='9') num++;
        }
        cout<<num<<endl;
        for(int i=0;i<s.size();i++)
        {
            if(s[i]>'0'&&s[i]<='9')
            {
                cout<<s[i];
                for(int j=1;j<=s.size()-i-1;j++)
                cout<<0;
                cout<<' ';
            }
        }
        cout<<endl;
    }
    return 0; 
}