贪心 Codeforces Round #300 A Cutting Banner

题目传送门

/*
    贪心水题：首先，最少的个数为n最大的一位数字mx，因为需要用1累加得到mx，
            接下来mx次循环，若是0，输出0；若是1，输出1，s[j]--；
    注意：之前的0的要忽略
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
using namespace std;

const int MAXN = 1e4 + 10;
const int INF = 0x3f3f3f3f;

int main(void)        //Codeforces Round #300 A Cutting Banner
{
    //freopen ("B.in", "r", stdin);

    char s[10];
    while (scanf ("%s", &s) == 1)
    {
        int len = strlen (s);
        int mx = -1;
        for (int i=0; i<len; ++i)    mx = max (mx, s[i] - '0');

        printf ("%d
", mx);
        for (int i=1; i<=mx; ++i)
        {
            bool ok = false;
            for (int j=0; j<len; ++j)
            {
                if (s[j] == '0')
                {
                    if (!ok)    continue;
                    else    printf ("0");
                }
                else
                {
                    ok = true;
                    printf ("1");    s[j]--;
                }
            }
            if (i < mx)    printf (" ");
        }
        puts ("");
    }

    return 0;
}