이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
// #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
using namespace std;
typedef long long ll;
typedef pair<int, int> pp;
#define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
#define per(i,r,l) for(int i = (r); i >= (l); i--)
#define rep(i,l,r) for(int i = (l); i < (r); i++)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define pb push_back
#define ss second
#define ff first
void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 1e6 + 5, lg = 21, inf = ll(1e9) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;}
int a[maxn], b[maxn], dpmx[maxn][2], dpmn[maxn][2];
int main(){
cin.tie(0) -> sync_with_stdio(0);
int n; cin >> n;
rep(i,0,n<<1) cin >> a[i];
rep(i,0,n<<1) cin >> b[i];
rep(i,0,2) dpmx[0][i] = dpmn[0][i] = i;
rep(i,1,n<<1){
rep(j,0,2) dpmn[i][j] = (n<<1)+1;
if(a[i] >= a[i-1]) dpmx[i][1] = max(dpmx[i][1], dpmx[i-1][1]+1), dpmn[i][1] = min(dpmn[i][1], dpmn[i-1][1]+1);
if(a[i] >= b[i-1]) dpmx[i][1] = max(dpmx[i][1], dpmx[i-1][0]+1), dpmn[i][1] = min(dpmn[i][1], dpmn[i-1][0]+1);
if(b[i] >= a[i-1]) dpmx[i][0] = max(dpmx[i][0], dpmx[i-1][1]), dpmn[i][0] = min(dpmn[i][0], dpmn[i-1][1]);
if(b[i] >= b[i-1]) dpmx[i][0] = max(dpmx[i][0], dpmx[i-1][0]), dpmn[i][0] = min(dpmn[i][0], dpmn[i-1][0]);
// er(i, dpmx[i][0], dpmx[i][1], dpmn[i][0], dpmn[i][1]);
}
int en = (n<<1)-1;
if((dpmx[en][0] < n || dpmn[en][0] > n) && (dpmx[en][1] < n || dpmn[en][1] > n)) return cout << "-1\n", 0;
bool cr = false;
if(dpmx[en][0] < n || dpmn[en][0] > n) cr = true;
string ans(n<<1,'A');
int cnt = 0;
per(i,en,0){
ans[i] = cr?'A':'B', cnt += cr;
if(i){
if(cnt + dpmx[i-1][cr] < n || cnt + dpmn[i-1][cr] > n || (cr?a[i]<a[i-1]:b[i]<b[i-1])) cr ^= 1;
}
}
cout << ans << '\n';
return 0;
}
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