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>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define plx pair<ll,int>
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(),x.end()
#define vi vector<int>
#define vl vector<ll>
#define vvi vector<vi>
using namespace std;
const ll inf=123456789,mxn=1e6+5;
int dpl[2][mxn]{0},dpr[2][mxn]{0},a[mxn],b[mxn];
int prl[2][mxn],prr[2][mxn];
int main(){
ios_base::sync_with_stdio(0);cin.tie(0);
int n;cin>>n;
for(int i=0;i<2*n;i++)cin>>a[i];
for(int i=0;i<2*n;i++)cin>>b[i];
dpl[0][0]=1;dpl[1][0]=0;prl[0][0]=prl[1][0]=-1;
for(int i=1;i<2*n;i++){
dpl[0][i]=dpl[1][i]=-1e9;
if(a[i]>=a[i-1]&&dpl[0][i]<dpl[0][i-1]+1)dpl[0][i]=dpl[0][i-1]+1,prl[0][i]=0;
if(a[i]>=b[i-1]&&dpl[0][i]<dpl[1][i-1]+1)dpl[0][i]=dpl[1][i-1]+1,prl[0][i]=1;
if(b[i]>=a[i-1]&&dpl[1][i]<dpl[0][i-1])dpl[1][i]=dpl[0][i-1],prl[1][i]=0;
if(b[i]>=b[i-1]&&dpl[1][i]<dpl[1][i-1])dpl[1][i]=dpl[1][i-1],prl[1][i]=1;
}dpr[0][2*n-1]=0,dpr[1][2*n-1]=1;prr[0][2*n-1]=prr[1][2*n-1]=-1;
for(int i=2*n-2;i>=0;i--){
dpr[0][i]=dpr[1][i]=-1e9;
if(a[i+1]>=a[i]&&dpr[0][i]<dpr[0][i+1])dpr[0][i]=dpr[0][i+1],prr[0][i]=0;
if(a[i+1]>=b[i]&&dpr[1][i]<dpr[0][i+1]+1)dpr[1][i]=dpr[0][i+1]+1,prr[1][i]=0;
if(b[i+1]>=a[i]&&dpr[0][i]<dpr[1][i+1])dpr[0][i]=dpr[1][i+1],prr[0][i]=1;
if(b[i+1]>=b[i]&&dpr[1][i]<dpr[1][i+1]+1)dpr[1][i]=dpr[1][i+1]+1,prr[1][i]=1;
}int ans[2*n]={0};int mem1=-1,mem2=-1,mem3=-1;
for(int i=0;i<2*n-1;i++){
if(a[i]<=a[i+1]&&dpl[0][i]+2*n-1-i-dpr[0][i+1]==n)mem1=i,mem2=0,mem3=0;
if(b[i]<=a[i+1]&&dpl[1][i]+2*n-1-i-dpr[0][i+1]==n)mem1=i,mem2=1,mem3=0;
if(a[i]<=b[i+1]&&dpl[0][i]+2*n-1-i-dpr[1][i+1]==n)mem1=i,mem2=0,mem3=1;
if(b[i]<=b[i+1]&&dpl[1][i]+2*n-1-i-dpr[1][i+1]==n)mem1=i,mem2=1,mem3=1;
if(mem1!=-1){
int cur=mem1,c=mem2;
while(cur>=0){
ans[cur]=c;
c=prl[c][cur];cur--;
}cur=mem1+1,c=mem3;
while(cur<=2*n-1){
ans[cur]=c;
c=prr[c][cur];cur++;
}
for(int i=0;i<2*n;i++){
if(ans[i])cout<<'B';
else cout<<'A';
}return 0;
}
}cout<<mem1;
}
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