A. Sum of Round Numbers
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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 11 to 99 (inclusive) are round.
For example, the following numbers are round: 40004000, 11, 99, 800800, 9090. The following numbers are not round: 110110, 707707, 222222, 10011001.
You are given a positive integer nn (1≤n≤1041≤n≤104). Represent the number nn as a sum of round numbers using the minimum number of summands (addends). In other words, you need to represent the given number nn as a sum of the least number of terms, each of which is a round number.
Input
The first line contains an integer tt (1≤t≤1041≤t≤104) — the number of test cases in the input. Then tt test cases follow.
Each test case is a line containing an integer nn (1≤n≤1041≤n≤104).
Output
Print tt answers to the test cases. Each answer must begin with an integer kk — the minimum number of summands. Next, kk terms must follow, each of which is a round number, and their sum is nn. The terms can be printed in any order. If there are several answers, print any of them.
Example
input
Copy
5
5009
7
9876
10000
10
output
Copy
2
5000 9
1
7
4
800 70 6 9000
1
10000
1
10
题意：输出各个位置的数字*权值，并输出位置，其实本题由于数字的范围的限制，权值很受限，如果数字很大，可以试试用字符串处理。
坑点：*前提是当前位置的数不是0，一但是0，直接排除！！！！！！！！！
标准程序：
#include<iostream>
using namespace std;
int w[5]={
1,10,100,1000,10000};
void Round(int n){
int num=n, k=0, i=0;
while(num){
if(num%10) k++;
num/=10;
}
printf("%d\n", k);
while(n){
if(n%10){
printf("%d ", (n%10)*w[i]);
}
n/=10;
i++;
}
printf("\n");
}
int main(){
int t,n;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
Round(n);
}
return 0;