Educational Codeforces Round 47 (Rated for Div. 2)--B. Minimum Ternary String
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.
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).
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
100210
001120
11222121
11112222
20
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 <stdio.h>
#include <iostream>
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;
}
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