Codeforces 600C Make Palindrome 【贪心 找字典序最小回文串】

C. Make Palindrome

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

题意：给定一个只包含小写字母的字符串，你可以修改任意位置的字符（变换为a-z中任一个），然后重新排列字符串。现在要求用最少次数的修改得到一个回文串，若有多种方案，输出字典序最小的方案。

思路：找出并记录所有字母的出现次数num[]，若num[i] & 1说明该字母不能对称放置，这时有两种方案——1，需要把其它字母改为当前字母；2，把该字母改为其它字母。为了得到最小字典序，我们贪心的选择将字典序大的字母改为字典序小的字母，按照这样的思维模拟即可。

注意：串长度len & 1时的处理。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (200000+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;char str[MAXN]; int num[30];bool mark[30];int main(){    Rs(str); CLR(num, 0);    int len = strlen(str);    for(int i = 0; i < len; i++)        num[str[i]-'a']++;    CLR(mark, false); int cnt = 0;    for(int i = 0; i < 26; i++)        if(num[i] & 1)            mark[i] = true, cnt++;    char mid; int use = 0;    if(len & 1)    {        cnt--;        for(int i = 0; i <= 25; i++)        {            if(mark[i])                use++;            if(use > cnt / 2)            {                mid = 'a' + i;                mark[i] = false;                break;            }        }    }    bool flag = false; use = 0;    for(int i = 0; i <= 25; i++)    {        if(mark[i])        {            if(flag)                num[i]--;            else                num[i]++;            use++;        }        if(use == cnt / 2)            flag = true;    }    int F = 0, B = len-1;    for(int i = 0; i < 26; i++)    {        for(int j = F; j < F + num[i] / 2; j++)            str[j] = 'a' + i;        F += num[i] / 2;        for(int j = B; j > B - num[i] / 2; j--)            str[j] = 'a' + i;        B -= num[i] / 2;    }    if(len & 1)        str[len / 2] = mid;    Ps(str);    return 0;}