Codeforces Round #300 Quasi Binary（DP）

Quasi Binary

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

The first line contains a single integer n (1 ≤ n ≤ 106).

Output

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.

Examples

input

9

output

9
1 1 1 1 1 1 1 1 1 

input

32

output

3
10 11 11 
【题意】给出一种定义数，这种数每一位不是0就是1，给你一个n，问你最少可以由多少定义数组成，分别列出这些定义数。
【分析】简单DP，先预处理出所有可能的数，然后背包。

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long LL;
const int N = 1e6+50;
int n;
int dp[N],vis[N],pre[N];
vector<int>vec,ans;
void init(){
    queue<int>q;
    q.push(1);
    vis[1]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vec.pb(u);
        int k = u*10+0;
        if(k<=1000000&&!vis[k]){
            q.push(k);
            vis[k]=1;
        }
        k = u*10+1;
        if(k<=1000000&&!vis[k]){
            q.push(k);
            vis[k]=1;
        }
    }
}
int main(){
    scanf("%d",&n);
    if(!n){
        printf("1
0
");
        return 0;
    }
    init();
    met(dp,inf);
    dp[0]=0;
    for(int i=1;i<=n;i++){
        for(int x : vec){
            if(i-x>=0&&dp[i-x]+1<dp[i]){
                dp[i]=dp[i-x]+1;
                pre[i]=i-x;
            }
        }
    }
    int k=n;
    while(k){
        int s=k-pre[k];
        k=pre[k];
        ans.pb(s);
    }
    printf("%d
",dp[n]);
    for(auto x:ans){
        printf("%d ",x);
    }printf("
");
    return 0;
}