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
100210
Output
001120
Input
11222121
Output
11112222
Input
20
Output
20
思路：很不錯的一道題目。1可以和0交換，也可以和2交換。那么1就要盡可能的往前。0和2不能交換，那么第一個2后面的那些0和2的相對位置是不會變的，因為1我們要讓它盡可能的往前，這樣才是最優(yōu)的。因此我們將第一個2前面的0放置在最前面，后面排上所有的1，剩下的0和2相對位置就不變了。
代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<stack> #include<map> #include<algorithm> #define inf 0x3f3f3f3f using namespace std; const int maxn=1e5+10; typedef long long ll; char s[maxn],str[maxn]; int main() {scanf("%s",&s);int n = strlen(s);int sum = 0； // 第一個2 前邊0 的個數(shù) int sum1 = 0; // 1 的個數(shù)int k = 0,flag = 0;for(int i=0;i<n;i++){if(s[i]=='0' && flag==0)sum++;else if(s[i]=='0' && flag==1)str[k++] = s[i];else if(s[i]=='1')sum1++;else if(s[i]=='2'){flag = 1;str[k++] = s[i];}}for(int i=0;i<sum;i++)printf("0");for(int i=0;i<sum1;i++)printf("1");printf("%s\n",str);return 0; }

努力加油a啊，(^o)/~

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