Codeforces1009B Minimum Ternary String（模拟+思维）题解

题目来源：

http://codeforces.com/problemset/problem/1009/B

题目描述：

题目描述

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" →→ "100210";
"010210" →→ "001210";
"010210" →→ "010120";
"010210" →→ "010201".

Note than you cannot swap "02" →→ "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 aa is lexicographically less than string bb (if strings aa and bb have the same length) if there exists some position ii (1≤i≤|a|1≤i≤|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 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

output

Copy

input

Copy

11222121

output

Copy

11112222

input

Copy

output

Copy

大致题意：

给一个数字串，只有数字0,1,2，你可以将相邻的两个数组调换，0和1,1,和2，但是0和2不能换，问经过任意次调换，可以形成最小的数字是什么，有前导零。。。

解题思路：

一开始想了半天，还分情况讨论，后来才知道，我们可以知道1是可以在任意的位置的而，2和0的相对位置不能变，所以我们只要在第一个2的前面加上所有的1就一定是最小的，记得考虑没有2的情况。。。。

代码：

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int main()
{
	string a,b;
	cin>>a;
	int len=a.length(),sum=0;
	for(int i=0;i<len;i++)
	{
		if(a[i]=='1')sum++;
		else b+=a[i];
	}
	len=b.length();
	for(int i=0;i<len;i++)
	{
		if(b[i]=='2'){
			for(int i=1;i<=sum;i++)
			cout<<"1";
			sum=0;
		}
		cout<<b[i];
	}
	for(int i=1;i<=sum;i++)
	cout<<"1";
	cout<<endl;
	return 0;
}