結果

問題 No.1051 PQ Permutation
ユーザー kiyoshi0205kiyoshi0205
提出日時 2020-05-08 22:55:33
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 144 ms / 2,000 ms
コード長 4,344 bytes
コンパイル時間 2,167 ms
コンパイル使用メモリ 189,504 KB
実行使用メモリ 15,744 KB
最終ジャッジ日時 2024-07-05 19:55:14
合計ジャッジ時間 5,819 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
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
sample AC * 3
other AC * 46
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>
// #include<ext/pb_ds/tag_and_trait.hpp>
// using namespace __gnu_pbds;
// #include<boost/multiprecision/cpp_int.hpp>
// namespace multiprecisioninteger = boost::multiprecision;
// using cint=multiprecisioninteger::cpp_int;
using namespace std;
using ll=long long;
#define double long double
using datas=pair<ll,ll>;
using ddatas=pair<double,double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
// using llset=tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;
#define For(i,a,b) for(i=a;i<(ll)b;++i)
#define bFor(i,b,a) for(i=b,--i;i>=(ll)a;--i)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N,0)
#define brep1(i,N) bFor(i,N,1)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define endl "\n"
#define eb emplace_back
#define print(v) cout<<v<<endl
#define printyes cout<<"Yes"<<endl
#define printno cout<<"No"<<endl
#define printYES cout<<"YES"<<endl
#define printNO cout<<"NO"<<endl
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
#define matoutput(v) do{for(auto outimat:v)output(outimat);}while(0)
const ll mod=1000000007;
// const ll mod=998244353;
const ll inf=1LL<<60;
const double PI = acos(-1);
const double eps = 1e-9;
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;}
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;}
void startupcpp(){
cin.tie(0);
ios::sync_with_stdio(false);
cout<<fixed<<setprecision(15);
}
double distance(ddatas x,ddatas y){
double a=x.first-y.first,b=x.second-y.second;
return sqrt(a*a+b*b);
}
ll modinv(ll a) {
ll b=mod,u=1,v=0,t;
while(b){
t=a/b;
a-=t*b; swap(a,b);
u-=t*v; swap(u,v);
}
return (u+mod)%mod;
}
ll moddevide(ll a,ll b){return (a*modinv(b))%mod;}
vec modncrlistp,modncrlistm;
ll modncr(ll n,ll r){
if(n<r)return 0;
ll i,size=modncrlistp.size();
if(size<=n){
modncrlistp.resize(n+1);
modncrlistm.resize(n+1);
if(!size){
modncrlistp[0]=modncrlistm[0]=1;
size++;
}
For(i,size,n+1){
modncrlistp[i]=modncrlistp[i-1]*i%mod;
modncrlistm[i]=modinv(modncrlistp[i]);
}
}
return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}
ll modpow(ll a,ll n){
ll res=1;
while(n>0){
if(n&1)res=res*a%mod;
a=a*a%mod;
n>>=1;
}
return res;
}
ll gcd(ll a,ll b){if(!b)return abs(a);return (a%b==0)?abs(b):gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll countdigits(ll n){
ll ans=0;
while(n){n/=10;ans++;}
return ans;
}
ll sumdigits(ll n){
ll ans=0;
while(n){ans+=n%10;n/=10;}
return ans;
}
int main(){
// startupcpp();
// int codeforces;cin>>codeforces;while(codeforces--){
ll i,f=0,N,A,B,pa=0,pb=0;
cin>>N>>A>>B;
vec v(N),ans(N);
rep(i,N)cin>>v[i];
if(!next_permutation(all(v))){
print(-1);
return 0;
}
rep(i,N){
if(v[i]==A)pa=i;
if(v[i]==B)pb=i;
}
if(pa<pb){
output(v);
return 0;
}
ans=v;
set<ll> se;
For(i,pb,N)se.insert(v[i]);
auto itr=se.upper_bound(B);
if(itr!=se.end()){
ans[pb]=*itr;
se.erase(itr);
if(A<B||A==ans[pb]){
For(i,pb+1,N){
ans[i]=*se.begin();
se.erase(se.begin());
}
}else{
itr=se.begin();
i=pb;
while(++i<N){
if(*itr==B)++itr;
ans[i]=*itr;
if(*itr==A){
ans[++i]=B;
}
++itr;
}
}
output(ans);
}else{
i=pb-1;
while(i>=0){
se.insert(v[i]);
if(B<v[i])f=1;
itr=se.upper_bound(v[i]);
if(itr!=se.end()&&(*itr!=B||f))break;
--i;
}
if(i==-1){
print(-1);
return 0;
}
if(*itr==B)++itr;
ans[i]=*itr;
se.erase(itr);
if(ans[i]==A){
itr=se.begin();
while(++i<N){
ans[i]=*itr;
++itr;
}
}else{
itr=se.begin();
while(++i<N){
if(*itr==B&&B<A)++itr;
ans[i]=*itr;
if(*itr==A&&B<A){
ans[++i]=B;
}
++itr;
}
}
output(ans);
}
}
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