Educational Codeforces Round 47 (Rated for Div. 2)--B. Minimum Ternary String

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

Note than you cannot swap "02" $\to$ "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 $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1\le i\le |a|$, where $|s|$ is the length of the string $s$) such that for every $j<i$

Input

The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and ${10}^5$ (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

题意：'0' 和 '1'可以互换 '1' 和 '2'可以互换，给我们一组数（只由0 ， 1，  2组成），通过使用这些互换任意次数（可能为0）来获得尽可能小的字典序该字符串 。

思路：由题意可知，1 可以在该字符串中任意穿梭，那么就可以先把1的数量加一下，最后在指定的位置输出即可

’2‘之前的’0‘一定可以换到最前面 ， 也记录一下’0‘的次数，最后在指定的位置输出

’2‘之后的’0‘不能移到前面去，按原样存到另一个字符串中， 最后再输出一下即可

代码：

#include 
#include 
using namespace std;
int main()
{
	string s1 , s2;
	while(cin >> s1)
	{
		int flag = 0;
		s2="";
		int a = 0 , b = 0;
		for(int i = 0 ; i < s1.length() ; i++)
		{
			if(s1[i] == '1')
			{
				a++;
			}
			if(flag == 0 && s1[i] == '0')
			{
				b++;
			}
			if(s1[i] =='2')
			{
				flag = 1;
				s2+='2';
			}
			if(flag == 1 && s1[i] == '0')
			{
				s2+='0';
			}
		}
		for(int i = 0 ; i < b ; i++)
		{
			printf("0");
		}
		for(int i = 0 ; i < a ; i++)
		{
			printf("1");
		}
		cout << s2 << endl;
	 } 
	return 0;
}