题意:
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.
题解:
因为分出来得数字只能是十进制,并且只可以用0和1组成,那么最少需要几个数,取决于给定得数得 从最低位到最高位中 数字最大得那一位。
其次我们对于那个数进行减操作即可,某位不为0就输出1,为0就输出0,直到全为0为止。
/*Keep on going Never give up*/
//#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define Accepted 0
#define AK main()
#define I_can signed
using namespace std;
const int maxn =2e5+10;
const int MaxN = 0x3f3f3f3f;
const int MinN = 0xc0c0c00c;
typedef long long ll;
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
using namespace std;
int sol(int x){
int res=0;
while(x){
res=max(res,x%10);
x/=10;
}
return res;
}
I_can AK{
ios::sync_with_stdio(false);
int n;
cin>>n;
int x=sol(n);
cout<<x<<endl;
string s=to_string(n);
for(int i=0;i<x;i++){
bool flag=false;
for(auto &j:s){
if(j>='1') flag=true;
if(j=='0'&&flag) cout<<"0";
else if(j!='0'){
j=j-1;
cout<<"1";
}
}
cout<<" ";
}
//cout<<s<<endl;
return 0;
}

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