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

朱雀 2021-09-15 17:26 217阅读 0赞

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" →→ "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

100210

output

Copy

001120

input

Copy

11222121

output

Copy

11112222

input

Copy

output

Copy

这场没打，很多人说卡了B,A了的代码也很毒

题意是0与1能交换位置，1与2能交换位置，求交换后最大的数

很明显0与2的相对位置不能变，1随便放，那么就没了？？？

先把0的个数保存，遇到2的时候就先输出前面的0，再输出所有的1，再输出2

#include <bits/stdc++.h>
    using namespace std;
    string str;
    int ze,on;
    int main()
    {
        cin>>str;
        for(int i = 0; i<str.size(); i++)
            if(str[i] == '1')
                on++;
        for(int i = 0; i<str.size(); i++)
            if(str[i] == '1')
                continue;
            else if(str[i] == '0')
                ze++;
            else
            {
                for(int j = 0; j<ze; j++)
                    cout<<0;
                for(int j = 0; j<on; j++)
                    cout<<1;
                cout<<2;
                ze = 0;
                on = 0;
            }
        for(int j = 0; j<ze; j++)
            cout<<0;
        for(int j = 0; j<on; j++)
            cout<<1;
        return 0;
    }