Codeforces Round #764 (Div. 3) D. Palindromes Coloring

Solution

균등하게 펠린드롬을 만들기 위해선 2개씩 짝지을 수 있는 알파벳들을 최대한 균등하게 k개의 슬롯에 분배하는 방법이다.

각 알파벳별 숫자에서 1을 버림한 짝수를 db에 더한다. 이 수는 페어로 펠린드롬을 만들 수 있는 개수이다. 단일로 들어가야하는 알파벳은 sg에 더한다.

각 슬롯에 들어갈 수 있는 짝수 알파벳 페어의 최소 값은 db / 2 / k * 2이다.
각 슬롯에 들어갈 수 있는 홀수 알파벳의 최소 값은 sg + (db % (k * 2)) >= k이다.

따라서 두 값을 더하면 균등하게 분배했을 때 만들 수 있는 컬러링된 펠린드롬의 최소 길이다.

#include <bits/stdc++.h>

using namespace std;

//Speed
#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define Sumfi cout.tie(NULL);

//Aliases
using ll= long long;
using lld= long double;
using ull= unsigned long long;

//Constants
const lld pi= 3.141592653589793238;
const ll INF= 1e18;
const ll mod=1e9+7;

//TypeDEf
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef unordered_map<ll,ll> umll;
typedef map<ll,ll> mll;

// Macros
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define fl(i,n) for(int i=0;i<n;i++)
#define rl(i,m,n) for(int i=n;i>=m;i--)
#define py cout<<"YES\n";
#define pm cout<<"-1\n";
#define pn cout<<"NO\n";
#define vr(v) v.begin(),v.end()
#define rv(v) v.end(),v.begin()

//Debug
#ifdef AbhinavAwasthi
#define debug(x) cerr << #x<<" "; cerr<<x<<" "; cerr << endl;
#else
#define debug(x);
#endif

// Operator overloads
template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

// Utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
void printarr(ll arr[], ll n){fl(i,n) cout << arr[i] << " ";cout << "\n";}
template<typename T>
void printvec(vector<T>v){ll n=v.size();fl(i,n)cout<<v[i]<<" ";cout<<"\n";}
template<typename T>
ll sumvec(vector<T>v){ll n=v.size();ll s=0;fl(i,n)s+=v[i];return s;}

// Mathematical functions
ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);}
ll lcm(ll a, ll b){return (a/gcd(a,b)*b);}
ll moduloMultiplication(ll a,ll b,ll mod){ll res = 0;a %= mod;while (b){if (b & 1)res = (res + a) % mod;b >>= 1;}return res;}
ll powermod(ll x, ll y, ll p){ll res = 1;x = x % p;if (x == 0) return 0;while (y > 0){if (y & 1)res = (res*x) % p;y = y>>1;x = (x*x) % p;}return res;}

//Graph-dfs
// bool gone[MN];
// vector<int> adj[MN];
// void dfs(int loc){
//     gone[loc]=true;
//     for(auto x:adj[loc])if(!gone[x])dfs(x);
// }

//To Remember
// vector<array<int, 2>> ranges(n);
//     cin >> ranges;

//     sort(all(ranges), [] (const array<int, 2>& X, const array<int, 2>& Y){
//     	return X[1] - X[0] < Y[1] - Y[0];
//     });

//Sorting
bool sorta(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b){return (a.second > b.second);}

//Bits
string decToBinary(int n){string s="";int i = 0;while (n > 0) {s =to_string(n % 2)+s;n = n / 2;i++;}return s;}
ll binaryToDecimal(string n){string num = n;ll dec_value = 0;int base = 1;int len = num.length();for(int i = len - 1; i >= 0; i--){if (num[i] == '1')dec_value += base;base = base * 2;}return dec_value;}

//Check
bool isPrime(ll n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
bool isPowerOfTwo(int n){if(n==0)return false;return (ceil(log2(n)) == floor(log2(n)));}
bool isPerfectSquare(ll x){if (x >= 0) {ll sr = sqrt(x);return (sr * sr == x);}return false;}

//Code by Abhinav Awasthi
//Language C++
//Practice->Success

//Code

const ll AND = INT_MAX - 1;
void sumfi()
{
    ll n,k,db(0),sg(0);
    string s;
    cin>>n>>k;
    cin>>s;
    int arr[26]{0,};
    fl(i,n) {
        arr[s[i] - 'a']++;
    }
    fl(i,26) {
        db += arr[i] & AND;
        sg += arr[i] & 1;
    }
    cout<<(((db/k)>>1)<<1) + ((sg + (db % (k<<1))) >= k)<<endl;
}
//Main
int main()
{
    Code By Sumfi
    ll t;
    cin>>t;
    fl(i,t)
    {
        sumfi();
    }
    return 0;
}
//End