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 maxn (int)(1e5+51)
#define all(x) x.begin(), x.end()
#define sz(x) (int) x.size()
#define endl '\n'
#define ll long long
#define pb push_back
#define ull unsigned long long
#define ii pair<int,int>
#define iii tuple<int,int,int>
#define inf 2000000001
#define mod 1000000007 //998244353
#define _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
template<typename X, typename Y> bool ckmin(X& x, const Y& y) { return (y < x) ? (x=y,1):0; }
template<typename X, typename Y> bool ckmax(X& x, const Y& y) { return (x < y) ? (x=y,1):0; }
const ll MOD = 1e9+9;
const ll P = 31;
vector<ll>pw,prHash;
int main(){_
int n;string s;cin>>n>>s;
pw.pb(1);
while(sz(pw)!=n+1)pw.pb((pw.back()*P)%MOD);
prHash.resize(n+1);
prHash[0] = 0;
if(n%2==0){
cout<<"NOT POSSIBLE"<<endl;
return 0;
}
for(int i=0;i<n;++i)prHash[i+1] = ((s[i]-'A'+1) + (P*prHash[i])%MOD)%MOD;
auto calcHash = [&](int l,int r){
return (MOD + prHash[r+1] - (prHash[l]*pw[r-l+1])%MOD)%MOD;
};
int ans = -1,anshash;
for(int i=0;i<n;++i){//try to remove every character from string
if(i<n/2){
ll left = calcHash(0,i-1);
ll right = calcHash(i+1,n/2);
ll rest = calcHash(n/2+1,n-1);
ll tot = ((left*pw[n/2-i])%MOD + right)%MOD;
if(tot==rest){
if(ans==-1)ans=i,anshash=tot;
else if(anshash!=tot){
cout<<"NOT UNIQUE"<<endl;
return 0;
}
}
}else if(i==n/2){
if(calcHash(0,n/2-1)==calcHash(n/2+1,n-1)){
if(ans==-1)ans=i,anshash=calcHash(0,n/2-1);
else if(anshash!=calcHash(0,n/2-1)){
cout<<"NOT UNIQUE"<<endl;
return 0;
}
}
}else{
ll rest = calcHash(0,n/2-1);
ll left = calcHash(n/2,i-1);
ll right = calcHash(i+1,n-1);
ll tot = ((left*pw[n-i-1])%MOD+right)%MOD;
if(tot==rest){
if(ans==-1)ans=i,anshash=tot;
else if(anshash!=tot){
cout<<"NOT UNIQUE"<<endl;
return 0;
}
}
}
}
if(ans==-1)cout<<"NOT POSSIBLE";
else cout<<(ans>=n/2?s.substr(0,n/2):s.substr(n/2+1,n/2));
}
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