# Quasi Binary
## 题面翻译
**题目描述**
给出一个数n，你需要将n写成若干个数的和，其中每个数的十进制表示中仅包含0和1。
问最少需要多少个数
**输入输出格式**
输入格式：
一行 一个数 n（1≤n≤10^6）
输出格式：
最少的数的个数，并给出一种方案。
## 题目描述
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer $ n $ . Represent it as a sum of minimum number of quasibinary numbers.
## 输入格式
The first line contains a single integer $ n $ ( $ 1<=n<=10^{6} $ ).
## 输出格式
In the first line print a single integer $ k $ — the minimum number of numbers in the representation of number $ n $ as a sum of quasibinary numbers.
In the second line print $ k $ numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal $ n $ . Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
## 样例 #1
### 样例输入 #1
```
9
```
### 样例输出 #1
```
9
1 1 1 1 1 1 1 1 1
```
## 样例 #2
### 样例输入 #2
```
32
```
### 样例输出 #2
```
3
10 11 11
```
#include<iostream> #include<cstdio> using namespace std; int Ans,num[100];//其实这里定义10就够了，因为最多9个数就一定可以构成n int main() { int n,Res; scanf("%d",&n); for(int bit=1;bit<=n;bit*=10)//将每一个res对应的数值在每一次for循环中分配给res个num数组中的数，比如2，分配给num[1]1和num{2]1,
比如3，对应其实是30，那么就分配给num[1]10,num[2]10,num[3]10,综上看来，num[1]是最大的,num[res]
是最小的，所以倒序输出，就是num[ans] { Res=(n/bit)%10;//每一位的数值 Ans=max(Ans,Res); for(int i=1;i<=Res;++i) num[i]+=bit; } printf("%d\n",Ans); for(;Ans;--Ans) printf("%d ",num[Ans]);//倒序输出，从小到大 putchar('\n'); return 0; }
一个数最少可以由它的数位中最大数的个数的01二位数组成。
心得:感觉这种思维题真的很难想，反正我是想不出哎