H. Palindromic Cut
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits.
He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar.
Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings.
Input
The first line contains an integer n (1 ≤ n ≤ 4·105) — the length of string s.
The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits.
Output
Print to the first line an integer k — minimum number of palindromes into which you can cut a given string.
Print to the second line k strings — the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length.
Examples
input
6
aabaac
output
2
aba aca
input
8
0rTrT022
output
1
02TrrT20
input
2
aA
output
2
a A
链接：http://codeforces.com/contest/883/problem/H
题意：给一个字符串，问最少能把该字符串拆分成多少个长度相等的回文串，输出回文串个数，以及长度相等的回文串。
思路：
0.若所有字符都出现了偶数次，那么直接输出1个回文串便可。
1.若一个字符 a 出现了奇数次，那么这个字符 a 肯定要构成一个新的回文串，把这个字符 a 记录下来，再记录这个字符 a 还能组成多少对 aa。
2.若落单的字符能与成对的字符恰好组合成 n/len 条长度为 len 的字符串，那么就直接输出。
3.若不能满足 2 ，则拆一对成双的字符，放到落单的字符里去，重复 2。
#include <bits/stdc++.h>
using namespace std;
int cnt[256];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin>>n;
string s;
cin>>s;
for(int i=0;i<s.length();i++)
{
cnt[s[i]]++;
}
vector<char> dan;
vector<char> shuan;
for(int i=0;i<=255;i++)
{
if(cnt[i])
{
if(cnt[i]%2)
{
cnt[i]--;
dan.push_back(char(i));
}
if(cnt[i])
{
while(cnt[i])
{
cnt[i]-=2;
shuan.push_back(char(i));
}
}
}
}
char ans[400005];
int j=0;
if(dan.size()==0)
{
cout<<1<<endl;
for(int i=0;i<n/2;i++)
{
ans[i] = ans[n-i-1] = shuan[j++];
}
ans[n]='\0';
return cout<<ans<<endl,0;
}
while(shuan.size()%dan.size())
{
dan.push_back(shuan.back());
dan.push_back(shuan.back());
shuan.pop_back();
}
int len = n/dan.size();
cout<<n/len<<endl;
for(auto ct:dan)
{
ans[len/2]=ct;
for(int i=0;i<len/2;i++)
{
ans[i] = ans[len-i-1] = shuan[j++];
}
ans[len]='\0';
cout<<ans<<" ";
}
return 0;
}转载请注明出处^ ^