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 + ;
 const int INF = 0x3f3f3f3f;

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

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

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

     return ;
 }