結果

問題 No.2663 Zero-Sum Submatrices
ユーザー karinohitokarinohito
提出日時 2024-03-25 18:18:26
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,373 bytes
コンパイル時間 4,611 ms
コンパイル使用メモリ 270,880 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-09-30 14:07:13
合計ジャッジ時間 5,385 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 31
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#include<atcoder/all>
using namespace atcoder;
using namespace std;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vvvvll = vector<vvvll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
using vvvvb = vector<vvvb>;
using vvvvvb = vector<vvvvb>;
#define all(A) A.begin(),A.end()
#define rep(i, n) for (ll i = 0; i < (ll) (n); i++)
template<class T>
bool chmax(T& p, T q, bool C = 1) {
if (C == 0 && p == q) {
return 1;
}
if (p < q) {
p = q;
return 1;
}
else {
return 0;
}
}
template<class T>
bool chmin(T& p, T q, bool C = 1) {
if (C == 0 && p == q) {
return 1;
}
if (p > q) {
p = q;
return 1;
}
else {
return 0;
}
}
ll modPow(long long a, long long n, long long p) {
if (n == 0) return 1; // 0
if (n == 1) return a % p;
if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;
long long t = modPow(a, n / 2, p);
return (t * t) % p;
}
ll gcd(ll(a), ll(b)) {
if (a == 0)return b;
if (b == 0)return a;
ll c = a;
while (a % b != 0) {
c = a % b;
a = b;
b = c;
}
return b;
}
ll sqrtz(ll N) {
ll L = 0;
ll R = sqrt(N) + 10000;
while (abs(R - L) > 1) {
ll mid = (R + L) / 2;
if (mid * mid <= N)L = mid;
else R = mid;
}
return L;
}
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vm>;
using vvvm = vector<vvm>;
vector<mint> fact, factinv, inv, factK;
const ll mod = 998244353;
void prenCkModp(ll n) {
//factK.resize(4*n+5);
fact.resize(n + 5);
factinv.resize(n + 5);
inv.resize(n + 5);
fact[0] = fact[1] = 1;
factinv[0] = factinv[1] = 1;
inv[1] = 1;
for (ll i = 2; i < n + 5; i++) {
fact[i] = (fact[i - 1] * i);
inv[i] = (mod - ((inv[mod % i] * (mod / i))));
factinv[i] = (factinv[i - 1] * inv[i]);
}
//factK[0]=1;
//for(ll i=1;i<4*n+5;i++){
// factK[i]=factK[i-1]*mint(K-i+1);
// //K*(K-1)*...*(K-i+1);
//}
}
mint nCk(ll n, ll k) {
if (n < k || k < 0) return 0;
return (fact[n] * ((factinv[k] * factinv[n - k])));
}
//mint nCkK(ll n,ll k){
// if(K<n||K-n<k)return 0;
// mint res=factK[n+k];
// res*=factK[n].inv();
// res*=factinv[k];
// return res;
//}
bool DEB=0;
bool is_biparate(vvll &G){
ll N=G.size();
vb seen(N,0);
vll P(N,-1);
rep(i,N){
if(seen[i])continue;
queue<ll> Q;
Q.push(i);
P[i]=0;
while(!Q.empty()){
ll p=Q.front();
Q.pop();
if(seen[p])continue;
seen[p]=1;
for(auto v:G[p]){
if(P[v]==-1){
P[v]=1-P[p];
}
if(P[v]==P[p])return 0;
if(!seen[v])Q.push(v);
}
}
}
return 1;
}
int main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
ll N,M;
cin>>N>>M;
ll cnt=0;
vvll AN(1<<N,vll(1<<M));
rep(i,1<<N){
rep(j,1<<M){
AN[i][j]=cnt;
cnt++;
}
if(i%2==1)reverse(all(AN[i]));
rep(j,1<<M){
cout<<AN[i][j]<<" \n"[j==(1<<M)-1];
}
}
}
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