Codeforces Round #640 (Div. 4)A. Sum of Round Numbers

题目链接:https://Codeforces/contest/1352/problem/A

problem description

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

input
5
5009
7
9876
10000
10
output
2
5000 9
1
7
4
800 70 6 9000
1
10000
1
10

1、先计算出输入数字不为0的位数
2、依次输出以该数字开头的整数

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
    int t,n,k,ans,m,temp;
    cin>>t;
    while(t--)
    {
        k=10;
        ans=0;
        m=1;
        cin>>n;
        while(n/(m*10))
            m=m*10;
        temp=n;
        while(temp)
        {
            if(temp%k!=0)
                ans++;
            temp=temp/10;
        }
        cout<<ans<<endl;
        k=10;
        while(m)
        {
            if(n/m)
                cout<<n/m*m<<' ';
            n=n%m;
            m=m/10;
        }
        cout<<endl;
    }
    return 0;
}