codeforces 538B

B. 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 
这道题的意思为输入一个数，然后分解成只有1和0组成的数的和，输出这些数的个数和这些数。

#include<iostream>
#define MAXN 10
using namespace std;
int number[MAXN];
int newnumber[MAXN];
int dit[MAXN][MAXN];
int main()
{
    memset(number, 0, MAXN);
    memset(newnumber, 0, MAXN);
    memset(dit, 0, MAXN);
    int n, i, sum = 0;
    cin >> n;
    for (i = MAXN - 1;i > 0;i--)
    {
        number[i] = n % 10;
        n /= 10;
    }
    for (i = MAXN - 1;i > 0;i--)
    {
        int j, temp;
        temp = number[i];
        for (j = 0;j < temp;j++)
            dit[j][i] = 1;
    }
    for(i=0;i<MAXN;i++)
        for (int j = 0;j < MAXN;j++)
        {
            newnumber[i] = dit[i][j] + newnumber[i] * 10;
        }
    for (i = 0;i < MAXN;i++)
        if (newnumber[i] != 0)
            sum++;
    cout << sum << endl;
    for (i = 0;i < sum;i++)
        cout << newnumber[i] << ' ';
    cout << endl;
    return 0;
}

实现过程略显复杂，用了三个数组，而且复杂度较高。