B. Minimum Ternary String （这个B有点狠）

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" $\to$
"010210" $\to$
"010210" $\to$
"010210" $\to$

Note than you cannot swap "02" $\to$

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

String $a$

Input

The first line of the input contains the string $s$

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

output

Copy

input

Copy

11222121

output

Copy

11112222

input

Copy

output

Copy

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 }

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