CF1009B Minimum Ternary String 思维

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

20

output

Copy

20

题意：给你一串由0，1，2组成的字符串，除了0和2不能交换，其他任意字符可以两两交换，问能交换得到的最小字典序字符串是啥？

分析：首先记录所有1的个数，然后再遍历一次数组，记录0的个数，遇到2的时候判断，如果是第一次遇到2则先输出记录的0的个数，然后再输出1的个数，接着输出个2，后面遇到2的时候就不要再输出1（因为1和2可以交换，所以1都会被换到第一次输出去），

最后到结尾的时候输出0

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn], vis[maxn];
int main() {
    string s;
    while( cin >> s ) {
        ll a = 0, b = 0, c = 0;
        for( ll i = 0; i < s.length(); i ++ ) {
            if( s[i] == '1' ) {
                b ++;
            }
        }
        for( ll i = 0; i < s.length(); i ++ ) {
            if( s[i] == '0' ) {
                a ++;
            } else if( s[i] == '2' ) {
                c ++;
                for( ll j = 0; j < a; j ++ ) {
                    cout << 0;
                }
                if( c == 1 ) {
                    for( ll j = 0; j < b; j ++ ) {
                        cout << 1;
                    }
                }
                cout << 2;
                a = 0, b = 0;
            }
            if( i == s.length()-1 ) {
                for( ll j = 0; j < a; j ++ ) {
                    cout << 0;
                }
                if( c == 0 ) {
                    for( ll j = 0; j < b; j ++ ) {
                        cout << 1;
                    }
                }
            }
        }
        cout << endl;
    }
    return 0;
}