贪心 Codeforces Round #300 A Cutting Banner

题目传送门

 1 /*
 2     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，
 3             接下来mx次循环，若是0，输出0；若是1，输出1，s[j]--；
 4     注意：之前的0的要忽略
 5 */
 6 #include <cstdio>
 7 #include <iostream>
 8 #include <cstring>
 9 #include <string>
10 #include <algorithm>
11 #include <cmath>
12 #include <set>
13 #include <map>
14 using namespace std;
15
16 const int MAXN = 1e4 + 10;
17 const int INF = 0x3f3f3f3f;
18
19 int main(void)        //Codeforces Round #300 A Cutting Banner
20 {
21     //freopen ("B.in", "r", stdin);
22
23     char s[10];
24     while (scanf ("%s", &s) == 1)
25     {
26         int len = strlen (s);
27         int mx = -1;
28         for (int i=0; i<len; ++i)    mx = max (mx, s[i] - ‘0‘);
29
30         printf ("%d\n", mx);
31         for (int i=1; i<=mx; ++i)
32         {
33             bool ok = false;
34             for (int j=0; j<len; ++j)
35             {
36                 if (s[j] == ‘0‘)
37                 {
38                     if (!ok)    continue;
39                     else    printf ("0");
40                 }
41                 else
42                 {
43                     ok = true;
44                     printf ("1");    s[j]--;
45                 }
46             }
47             if (i < mx)    printf (" ");
48         }
49         puts ("");
50     }
51
52     return 0;
53 }