B. Minimum Ternary String （这个B有点狠）

Posted 2020-12-04 Fitz

B. Minimum Ternary String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a ternary string (it is a string which consists only of characters ‘0‘, ‘1‘ and ‘2‘).

You can swap any two adjacent (consecutive) characters ‘0‘ and ‘1‘ (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters ‘1‘ and ‘2‘ (i.e. replace "12" with "21" or vice versa).

For example, for string "010210" we can perform the following moves:

• "010210" $\to \to "100210";$
• "010210" $\to \to "001210";$
• "010210" $\to \to "010120";$
• "010210" $\to \to "010201".$

Note than you cannot swap "02" $\to \to "20"\; and\; vice\; versa.\; You\; cannot\; perform\; any\; other\; operations\; with\; the\; given\; string\; excluding\; described\; above.$

You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).

String $\mathrm{aa is\; lexicographically\; less\; than\; string bb (if\; strings aa and bb have\; the\; same\; length)\; if\; there\; exists\; some\; position ii (1\le i\le |a|1\le i\le |a|,\; where|s||s| is\; the\; length\; of\; the\; string ss)\; such\; that\; for\; every j<ij<i holds aj=bjaj=bj,\; and ai<biai<bi.}$

Input

The first line of the input contains the string $\mathrm{ss consisting\; only\; of\; characters\; ‘0‘,\; ‘1‘\; and\; ‘2‘,\; its\; length\; is\; between 11 and 105105 (inclusive).}$

Output

Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).

Examples

input

Copy

100210

output

Copy

001120

input

Copy

11222121

output

Copy

11112222

input

Copy

20

output

Copy

20

   这场CF abcd题  我就觉得B题最难  

  这个题目卡了我好久好久  ，全是写BUG

　　代码有点毒瘤   

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 const int INF = 0x3fffffff;
 5 typedef long long LL;
 6 string s;
 7 int sum0[maxn], sum1[maxn], sum2[maxn], vis[maxn];
 8 vector<int>a;
 9 int main() {
10     cin >> s;
11     if (s[0] == ‘0‘) sum0[1] = 1;
12     if (s[0] == ‘1‘) sum1[1] = 1;
13     if (s[0] == ‘2‘) sum2[1] = 1;
14     for (int i = 1 ; i < s.size() ; i++) {
15         if (s[i] == ‘0‘) sum0[i + 1] = sum0[i] + 1, sum1[i + 1] = sum1[i], sum2[i + 1] = sum2[i];
16         if (s[i] == ‘1‘) sum1[i + 1] = sum1[i] + 1, sum0[i + 1] = sum0[i], sum2[i + 1] = sum2[i];
17         if (s[i] == ‘2‘) sum2[i + 1] = sum2[i] + 1, sum0[i + 1] = sum0[i], sum1[i + 1] = sum1[i];
18     }
19     int cnt = 0;
20     int len = s.size();
21     for (int i = 0 ; i < len  ; i++) {
22         if (s[i] == ‘2‘) {
23             for (int j = 0 ; j < sum0[i]; j++) a.push_back(0);
24             for (int j = 0 ; j < sum1[len] ; j++) a.push_back(1);
25             for (int j = i ; j < s.size(); j++) {
26                 if (s[j] == ‘1‘) continue;
27                 if (s[j] == ‘2‘) a.push_back(2);
28                 if (s[j] == ‘0‘)a.push_back(0);
29             }
30             cnt = 1;
31             break;
32         }
33     }
34     if (cnt == 0) {
35         for (int i = 0 ; i < sum0[len] ; i++) printf("0");
36         for (int i = 0 ; i < sum1[len] ; i++) printf("1");
37     } else {
38         for (int i = 0 ; i < a.size() ; i++)  printf("%d", a[i]);
39     }
40     return 0;
41 }