結果
| 問題 |
No.1974 2x2 Flipper
|
| コンテスト | |
| ユーザー |
milkcoffee
|
| 提出日時 | 2022-06-10 22:22:21 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 83 ms / 2,000 ms |
| コード長 | 5,234 bytes |
| コンパイル時間 | 4,732 ms |
| コンパイル使用メモリ | 265,364 KB |
| 最終ジャッジ日時 | 2025-01-29 20:00:22 |
|
ジャッジサーバーID (参考情報) |
judge2 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 25 |
ソースコード
#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
using namespace std;
/*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
*/
#define ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep_up(i, a, n) for (ll i = a; i < n; ++i)
#define rep_down(i, a, n) for (ll i = a; i >= n; --i)
#define P pair<ll, ll>
#define pb push_back
#define bit_count(x) __builtin_popcountll(x)
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) a / gcd(a,b) * b
#define endl "\n"
#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define vvvvll vector< vector <vector <vector<ll> > > >
#define vvvll vector< vector< vector<ll> > >
#define vvll vector< vector<ll> >
#define vll vector<ll>
#define pqll priority_queue<ll>
#define pqllg priority_queue<ll, vector<ll>, greater<ll>>
template<class T> inline void vin(vector<T>& v) { rep(i, v.size()) cin >> v.at(i); }
template <class T>
using V = vector<T>;
constexpr ll INF = (1ll << 60);
constexpr ll mod = 1000000007;
//constexpr ll mod = 998244353;
constexpr double pi = 3.14159265358979323846;
template <typename T>
inline bool chmax(T &a, T b)
{
if (a < b)
{
a = b;
return 1;
}
return 0;
}
template <typename T>
inline bool chmin(T &a, T b)
{
if (a > b)
{
a = b;
return 1;
}
return 0;
}
template <typename T>
void pt(T val)
{
cout << val << "\n";
}
template <typename T>
void pt_vll(vector<T> &v)
{
ll vs = v.size();
rep(i, vs)
{
cout << v[i];
if (i == vs - 1)
cout << "\n";
else
cout << " ";
}
}
ll mypow(ll a, ll n)
{
ll ret = 1;
if (n == 0)
return 1;
if (a == 0)
return 0;
rep(i, n)
{
if (ret > (ll)(9e18 + 10) / a)
return -1;
ret *= a;
}
return ret;
}
long long modpow(long long a, long long n, long long mod)
{
long long res = 1;
while (n > 0)
{
if (n & 1)
res = res * a % mod;
a = a * a % mod;
n >>= 1;
}
return res;
}
long long modinv(long long a, long long m)
{
long long b = m, u = 1, v = 0;
while (b)
{
long long t = a / b;
a -= t * b;
swap(a, b);
u -= t * v;
swap(u, v);
}
u %= m;
if (u < 0)
u += m;
return u;
}
const int MAX = 3010000;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit()
{
fac[0] = fac[1] = 1;
finv[0] = finv[1] = 1;
inv[1] = 1;
for (ll i = 2; i < MAX; i++)
{
fac[i] = fac[i - 1] * i % mod;
inv[i] = mod - inv[mod % i] * (mod / i) % mod;
finv[i] = finv[i - 1] * inv[i] % mod;
}
}
// 二項係数計算
long long COM(ll n, ll k)
{
if (n < k)
return 0;
if (n < 0 || k < 0)
return 0;
return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}
vector<ll> enum_div(ll n)
{ //約数全列挙
vector<ll> ret;
for (ll i = 1; i * i <= n; ++i)
{
if (n % i == 0)
{
ret.push_back(i);
if (i * i != n)
{
ret.push_back(n / i);
}
}
}
return ret;
}
void make_prime(vector<ll> &ret, ll n)
{ //素因数分解
ll x = n;
for (ll i = 2; i * i <= x; i++)
{
while (n % i == 0)
{
n /= i;
ret.push_back(i);
}
}
if (n != 1)
{
ret.push_back(n);
}
return;
}
vector<bool> prime(1000010, true);
vector<ll> pri(1000010);
vector<bool> isprime(int N)
{ //素数判定
if (N >= 0)
prime[0] = false;
if (N >= 1)
prime[1] = false;
for (ll i = 2; i * i <= N; i++)
{
if (!prime[i])
{
continue;
}
for (ll j = i * i; j <= N; j += i)
{
if (prime[j])
pri[j] = i;
prime[j] = false;
}
}
return prime;
}
map<ll,ll> compression(vector<ll> v){
map<ll,ll> ret;
ll cnt = 0;
sort(v.begin(),v.end());
for(ll i=0; i<v.size(); i++){
if(!ret.count(v[i])){
ret[v[i]] = cnt;
cnt++;
}
}
return ret;
}
//素因数分解: void make_prime(vector<ll> &ret, ll n)
//素数判定 : vector<bool> isprime(int N)
//約数全列挙: vector<ll> enum_div (ll n)
void solve(){
ll n, m, k, cnt = 0, sum = 0, ans = 0;
cin>>n>>m;
vvll a(n,vll(m));
rep(i,n){
rep(j,m){
a[i][j]=1;
if(i==n-1&&n%2==1) a[i][j]=0;
if(j==m-1&&m%2==1) a[i][j]=0;
}
}
ans=(n/2)*2*(m/2)*2;
if(n*m%2==1&&n>=2&&m>=2){
ans=n*m-max(n,m);
rep(i,n){
rep(j,m){
if(i==j) a[i][j]=0;
else if((n<m&&i==n-1&&j>=n-1)||(n>=m&&j==m-1&&i>=m-1)) a[i][j]=0;
else a[i][j]=1;
}
}
}
pt(ans);
rep(i,n){
pt_vll(a[i]);
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
//cout << fixed << setprecision(16);
//ll T;
//cin>>T;
//rep(ca,T)
solve();
}
milkcoffee