标签：pen sam std display ssis splay bytes deb 字典
http://codeforces.com/problemset/problem/1009/B
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:
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
代码如下：
#include <bits/stdc++.h> using namespace std; int main(){ string str; cin>>str; int len=str.size(); int pos; int flag=0; int num0=0,num1=0; for(int i=0;i<len;i++){ if(str[i]==‘2‘&&flag==0){ flag=1; pos=i; } if(str[i]==‘1‘) num1++; if(str[i]==‘0‘&&flag==0) num0++; } for(int i=0;i<num0;i++){ cout<<"0"; } for(int i=0;i<num1;i++){ cout<<"1"; } if(flag==1){ for(int i=pos;i<len;i++){ if(str[i]==‘1‘) continue; cout<<str[i]; } } cout<<endl; return 0; }
codeforces ~ 1009 B Minimum Ternary String(超级恶心的思维题
标签：pen sam std display ssis splay bytes deb 字典
原文地址：https://www.cnblogs.com/buerdepepeqi/p/9311722.html
