not ade row atom times possibly and sidebar form

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

ss consisting only of characters ‘0‘, ‘1‘ and ‘2‘, its length is between $\mathrm{11\; and\; 105105\; (inclusive).}$

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

這場CF abcd題 我就覺得B題最難

這個題目卡了我好久好久 ，全是寫BUG

　　代碼有點毒瘤

 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 }

B. Minimum Ternary String （這個B有點狠）