zoukankan      html  css  js  c++  java
  • codeforces Educational Codeforces Round 2 C Make Palindrome

    C. Make Palindrome

     
     

    A string is called palindrome if it reads the same from left to right and from right to left. For example "kazak", "oo", "r" and "mikhailrubinchikkihcniburliahkim" are palindroms, but strings "abb" and "ij" are not.

    You are given string s consisting of lowercase Latin letters. At once you can choose any position in the string and change letter in that position to any other lowercase letter. So after each changing the length of the string doesn't change. At first you can change some letters in s. Then you can permute the order of letters as you want. Permutation doesn't count as changes.

    You should obtain palindrome with the minimal number of changes. If there are several ways to do that you should get the lexicographically (alphabetically) smallest palindrome. So firstly you should minimize the number of changes and then minimize the palindrome lexicographically.

    Input

    The only line contains string s (1 ≤ |s| ≤ 2·105) consisting of only lowercase Latin letters.

    Output

    Print the lexicographically smallest palindrome that can be obtained with the minimal number of changes.

    Sample test(s)
    Input
    aabc
    Output
    abba
    Input
    aabcd
    Output
    abcba
     1 #include<cstdio>
     2 #include<iostream>
     3 #include<vector>
     4 #include<cmath>
     5 #include<queue>
     6 #include<string>
     7 #include<map>
     8 #include<cstring>
     9 #include<algorithm>
    10 using namespace std;
    11 typedef long long ll;
    12 typedef unsigned long long ull;
    13 const int maxn=2e5+5;
    14 int cnt[27];
    15 int main()
    16 {
    17     char s[maxn];
    18     int flag=-1;
    19     scanf("%s",s);
    20     cnt[26]=0;
    21     int len=strlen(s);
    22     for(int i=0;i<len;i++)cnt[s[i]-'a']++;
    23     for(int i=0;i<26;i++)
    24     {
    25         if(cnt[i]&1)
    26             for(int j=25;j>i;j--)
    27                 if(cnt[j]&1)
    28                 {
    29                     cnt[i]++;
    30                     cnt[j]--;
    31                     break;
    32                 }
    33     }
    34     for(int i=0;i<26;i++)
    35     {
    36         if(cnt[i]&1)flag=i;
    37         for(int j=0;j<cnt[i]/2;j++)
    38             printf("%c",i+'a');
    39     }
    40     if(flag>=0)
    41         printf("%c",flag+'a');
    42     for(int i=25;i>=0;i--)
    43             for(int j=0;j<cnt[i]/2;j++)
    44                 printf("%c",i+'a');
    45     puts("");
    46     return 0;
    47 }
  • 相关阅读:
    单例模式
    eclipse部署web项目至本地的tomcat但在webapps中找不到
    使用 google gson 转换Timestamp为JSON字符串
    前端JS对后台传递的timestamp的转换
    2018第15周总结
    Tomcat的最大并发数
    平台对接的另外一种模式
    系统间数据交换的5种方式
    Spring Boot条件注解
    Spring Data JPA 和MyBatis比较
  • 原文地址:https://www.cnblogs.com/homura/p/5002164.html
Copyright © 2011-2022 走看看