Submission #4612658


Source Code Expand

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define tr(container, it) \
    for (auto it = container.begin(); it != container.end(); it++)
#define scontains(c, x) ((c).find(x) != (c).end())   //O(log n)
#define contains(c, x) (find((c).begin(),(c).end(),x) != (c).end()) //O(n)
#define ill(_x)  ll _x;scanf("%lld",&_x);
#define idb(_x)  double _x;scanf("%lf",&_x);
#define pll pair<ll,ll>
#define mll map<ll,ll>
#define vll vector<ll>
#define sll set<ll>
#define vs vector<string>
#define in0(x, a, b)((x)>=a && (x)<=b    )
#define in1(x, a, b)((x)>a && (x)<b)
#define  rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define  _for(i, end) for (__typeof(end) i = 0; i < (end); i += 1)
#define all(x) (x).begin(),(x).end()
//#define len(array)  (sizeof(array)/sizeof((array)[0]))
#define endl '\n'
#define int ll

const double PI = 3.14159265358979323846;
const int INF = 0x3f3f3f3f;
const int LLINF = 1000000000000000005LL;;
const int MOD = (int) (1e9) + 7;
const int MX = 777777 + 5;
const int mod = 1777777777;
const double EPS = 1e-10;
const ll MXP = 8999251775770639633;
const int _500k = 500005;
const int _1m = 1000005;
const int _200k = 200005;
const int _20k = 20005;
const int _100k = 100005;
const int _10k = 10005;
int readint(){int x;cin >> x;return x;}

template <typename T>
ostream& operator << (ostream& os, const vector<T>& v){
    for(auto a : v)os << a << " ";
    return os;
}
template <typename T>
ostream& operator << (ostream& os, const set<T>& v){
    for(auto a : v)os << a << " ";
    return os;
}


inline int len(string s){return s.size();}
inline int len(int* A){return (sizeof(A))/sizeof(A[0]);}
template <typename T>inline int len(vector<T> v){return v.size();}
template <typename T>inline int len(set<T> v){return v.size();}
template <typename T,typename S>inline int len(map<T,S> v){return v.size();}

int fac[MX+5];
int power_mod(int x, int y, int p)
{
    if(y<0)return 0;
    int res = 1;

    x = x % p;

    while (y)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}

int power(int x, int y)
{
    if(y<0)return 0;
    int res = 1;
    while (y)
    {
        if (y & 1)
            res = (res*x);
        y = y>>1;
        x *=x ;
    }
    return res;
}


// Returns n^(-1) mod p
int modInverse(int n, int p)
{
    return power_mod(n, p-2, p);
}

int C(int n, int r, int p)
{
    if (r==0)
        return 1;

    return (fac[n]* modInverse(fac[r], p) % p *
            modInverse(fac[n-r], p) % p) % p;
}

void init(int mod){
    fac[0] = 1;
    for (int i=1 ; i<=MX; i++)
        fac[i] = fac[i-1]*i%mod;
}

int CC(int a, int b){
    int res = 1;
    rep(i,a+1,a-b+1)res *= i,res/=a-i+1;
//    rep(i,1,b+1)res /= i;
    return res;
}

bool equal(double a, double b) {
    return std::fabs(a - b) < std::numeric_limits<double>::epsilon();
}
struct UnionFind
{
    // par[i]:データiが属する木の親の番号。i == par[i]のとき、データiは木の根ノードである
    vector<int> par;
    // sizes[i]:根ノードiの木に含まれるデータの数。iが根ノードでない場合は無意味な値となる
    vector<int> sizes;

    UnionFind(int n) : par(n), sizes(n, 1) {
        // 最初は全てのデータiがグループiに存在するものとして初期化
        rep(i,0,n) par[i] = i;
    }

    // データxが属する木の根を得る
    int find(int x) {
        if (x == par[x]) return x;
        return par[x] = find(par[x]);  // 根を張り替えながら再帰的に根ノードを探す
    }

    // 2つのデータx, yが属する木をマージする
    void unite(int x, int y) {
        // データの根ノードを得る
        x = find(x);
        y = find(y);

        // 既に同じ木に属しているならマージしない
        if (x == y) return;

        // xの木がyの木より大きくなるようにする
        if (sizes[x] < sizes[y]) swap(x, y);

        // xがyの親になるように連結する
        par[y] = x;
        sizes[x] += sizes[y];
        // sizes[y] = 0;  // sizes[y]は無意味な値となるので0を入れておいてもよい
    }

    // 2つのデータx, yが属する木が同じならtrueを返す
    bool same(int x, int y) {
        return find(x) == find(y);
    }

    // データxが含まれる木の大きさを返す
    int size(int x) {
        return sizes[find(x)];
    }
};


//fastest
struct cpmFunctor{
    inline bool operator()(const pair<int,int> &p1, const pair<int,int> & p2){
        return p1.first < p2.first || (p1.first == p2.first && p1.second < p2.second);
    }
};

bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;
    if(n==2 || n==3)return true;
    if(n%2==0 || n%3==0)return false;
    // Check from 2 to n-1
    for (int i = 5; i < n; i+=6)
        if (n % i == 0)
            return false;
    for (int i = 7; i < n; i+=6)
        if (n % i == 0)
            return false;

    return true;
}

int cnt(string s,int f){
    int l=INF,r= -1,cnt = 0;
    _for(i,s.size()-f+1){
        if(s[i] == '-' ^ in0(i,l,r))l = max(r+1, (ll)i), r = i + f - 1,cnt ++;
    }
    rep(i,s.size()-f+1,s.size())if(s[i] == '-' ^ in0(i,l,r))return -1;
    return cnt;
}



int lcm(int a, int b){
    return a / __gcd(a,b) * b;
}

int m[35];
void _(){
    int h,w;
    cin >> h >> w;
    int cnt0 = 0, cnt1 = 0,cnt2 = 0,cnt3 =0;
    char t;
    _for(i,h*w){
        cin >> t;
        m[t-'a'] ++;
    }
    _for(i,26)cnt0 += (m[i]%4 == 0),cnt1 += (m[i]%4 == 1),cnt2 += (m[i]%4 == 2),cnt3 += (m[i]%4 == 3);
    if((h&1) && (w&1)){
        if(cnt1 == 1 && cnt3 ==0 && cnt2 <= h+w-2 || (cnt1 == 0 && cnt3 ==1 && cnt2 <= h+w-3))cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    else if((h&1) && !(w&1)){
        if(cnt2 <= w/2 && !cnt1 && !cnt3)cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    else if((w&1) && !(h&1)){
        if(cnt2 <= h/2 && !cnt1 && !cnt3)cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    else{
        if(!cnt1 && !cnt2 && !cnt3)cout << "Yes" << endl;
        else cout << "No" << endl;
    }

}

#undef int
int main() {
#if __MINGW32__
    freopen("C:\\Users\\hiroo\\CLionProjects\\competitive\\Input.txt", "r", stdin);
    freopen("C:\\Users\\hiroo\\CLionProjects\\competitive\\Output.txt", "w", stdout);

#endif

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    _();

}

Submission Info

Submission Time
Task C - Palindromic Matrix
User Gosu_Hiroo
Language C++14 (GCC 5.4.1)
Score 0
Code Size 6801 Byte
Status WA
Exec Time 1 ms
Memory 256 KB

Judge Result

Set Name Sample All
Score / Max Score 0 / 0 0 / 400
Status
AC × 5
AC × 30
WA × 2
Set Name Test Cases
Sample 0_00.txt, 0_01.txt, 0_02.txt, 0_03.txt, 0_04.txt
All 0_00.txt, 0_01.txt, 0_02.txt, 0_03.txt, 0_04.txt, 1_00.txt, 1_01.txt, 1_02.txt, 1_03.txt, 1_04.txt, 1_05.txt, 1_06.txt, 1_07.txt, 1_08.txt, 1_09.txt, 1_10.txt, 1_11.txt, 1_12.txt, 1_13.txt, 1_14.txt, 1_15.txt, 1_16.txt, 1_17.txt, 1_18.txt, 1_19.txt, 1_20.txt, 1_21.txt, 1_22.txt, 1_23.txt, 1_24.txt, 1_25.txt, 1_26.txt
Case Name Status Exec Time Memory
0_00.txt AC 1 ms 256 KB
0_01.txt AC 1 ms 256 KB
0_02.txt AC 1 ms 256 KB
0_03.txt AC 1 ms 256 KB
0_04.txt AC 1 ms 256 KB
1_00.txt AC 1 ms 256 KB
1_01.txt AC 1 ms 256 KB
1_02.txt AC 1 ms 256 KB
1_03.txt AC 1 ms 256 KB
1_04.txt AC 1 ms 256 KB
1_05.txt AC 1 ms 256 KB
1_06.txt AC 1 ms 256 KB
1_07.txt AC 1 ms 256 KB
1_08.txt AC 1 ms 256 KB
1_09.txt AC 1 ms 256 KB
1_10.txt AC 1 ms 256 KB
1_11.txt AC 1 ms 256 KB
1_12.txt AC 1 ms 256 KB
1_13.txt AC 1 ms 256 KB
1_14.txt AC 1 ms 256 KB
1_15.txt AC 1 ms 256 KB
1_16.txt AC 1 ms 256 KB
1_17.txt AC 1 ms 256 KB
1_18.txt AC 1 ms 256 KB
1_19.txt AC 1 ms 256 KB
1_20.txt AC 1 ms 256 KB
1_21.txt AC 1 ms 256 KB
1_22.txt WA 1 ms 256 KB
1_23.txt AC 1 ms 256 KB
1_24.txt AC 1 ms 256 KB
1_25.txt WA 1 ms 256 KB
1_26.txt AC 1 ms 256 KB