제출 #573518

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
573518		Sho10	앵무새 (IOI11_parrots)	C++17	0 / 100	3 ms	644 KiB

encoder

#include <bits/stdc++.h> //Andrei Alexandru a.k.a Sho
#include "encoder.h"
#include "encoderlib.h"
using ll=long long;
using ld=long double;
int const INF=1000000005;
ll const LINF=1000000000000000005;
ll const mod=6700417;
ld const PI=3.14159265359;
ll const MAX_N=3e5+5;
ld const EPS=0.00000001;
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define endl '\n'
#define sz(a) (int)a.size()
#define CODE_START  ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
map<ll,vector<ll>>viz;
string calc(ll nr){
string s="";
for(ll i=0;i<6;i++)
{
    if((1ll<<i)&nr){
        s+='1';
    }else s+='0';
}
reverse(s.begin(),s.end());
return s;
}
int turnback(string s){
ll ans=0,d=0;
for(ll i=s.size()-1;i>=0;i--)
{
    if(s[i]=='1'){
        ans+=(1ll<<d);
    }
    d++;
}
return ans;
}
void encode(int n,int a[]){
    ll val=0;
for(ll i1=0;i1<=3;i1++)
for(ll i2=i1;i2<=3;i2++)
for(ll i3=i2;i3<=3;i3++)
for(ll i4=i3;i4<=3;i4++)
for(ll i5=i4;i5<=3;i5++)
for(ll i6=i5;i6<=3;i6++)
for(ll i7=i6;i7<=3;i7++)
{
    for(ll i8=i7;i8<=3;i8++){
vector<ll>x;
x.pb(i1);
x.pb(i2);
x.pb(i3);
x.pb(i4);
x.pb(i5);
x.pb(i6);
x.pb(i7);
x.pb(i8);
viz[val]=x;
val++;
}
}
for(ll i=0;i<n;i++)
{
    vector<ll>x=viz[a[i]];
    string s=calc(i);
    for(auto it : x){
        if(it==0){
          //  continue;
            s+='0';
            s+='0';
        }else if(it==1){
        s+='0';
        s+='1';
        }else if(it==2){
        s+='1';
        s+='0';

    }else if(it==3){
    s+='1';
    s+='1';
    }
    send(turnback(s));
    s.pop_back();
    s.pop_back();
    s.pop_back();
}
}
}
/*
int32_t main(){
int n,a[25];
n=8;
a[0]=1;
a[1]=0;
a[2]=1;
a[3]=0;
a[4]=0;
a[5]=1;
a[6]=1;
a[7]=0;
encode(n,a);
}
*/

decoder

#include <bits/stdc++.h> //Andrei Alexandru a.k.a Sho
#include "decoder.h"
#include "decoderlib.h"
using ll=long long;
using ld=long double;
int const INF=1000000005;
ll const LINF=1000000000000000005;
ll const mod=6700417;
ld const PI=3.14159265359;
ll const MAX_N=3e5+5;
ld const EPS=0.00000001;
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define endl '\n'
#define sz(a) (int)a.size()
#define CODE_START  ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
map<vector<ll>,ll>viz;
string calc(ll nr){
string s="";
for(ll i=0;i<9;i++)
{
    if((1ll<<i)&nr){
        s+='1';
    }else s+='0';
}
reverse(s.begin(),s.end());
return s;
}
int turnback(string s){
ll ans=0,d=0;
for(ll i=s.size()-1;i>=0;i--)
{
    if(s[i]=='1'){
        ans+=(1ll<<d);
    }
    d++;
}
return ans;
}
void decode(int n,int m,int a[]){
        ll val=0;
for(ll i1=0;i1<=3;i1++)
for(ll i2=i1;i2<=3;i2++)
for(ll i3=i2;i3<=3;i3++)
for(ll i4=i3;i4<=3;i4++)
for(ll i5=i4;i5<=3;i5++)
for(ll i6=i5;i6<=3;i6++)
for(ll i7=i6;i7<=3;i7++)
{
    for(ll i8=i7;i8<=3;i8++){
vector<ll>x;
x.pb(i1);
x.pb(i2);
x.pb(i3);
x.pb(i4);
x.pb(i5);
x.pb(i6);
x.pb(i7);
x.pb(i8);
viz[x]=val;
val++;
}
}
vector<ll>g[65];
for(ll i=0;i<m;i++)
{
    string s=calc(a[i]);
    string x="";
    x+=s.back();
    s.pop_back();
    x+=s.back();
    s.pop_back();
    x+=s.back();
    s.pop_back();
    swap(x[0],x[2]);
    ll pos=turnback(s);
    ll val=turnback(x);
    g[pos].pb(val);
}
for(ll i=0;i<n;i++)
    sort(g[i].begin(),g[i].end());
    int ans[65];
    for(ll i=0;i<n;i++)
       output(viz[g[i]]);
}

컴파일 시 표준 에러 (stderr) 메시지

decoder.cpp: In function 'void decode(int, int, int*)':
decoder.cpp:85:1: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
   85 | for(ll i=0;i<n;i++)
      | ^~~
decoder.cpp:87:5: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
   87 |     int ans[65];
      |     ^~~
decoder.cpp:87:9: warning: unused variable 'ans' [-Wunused-variable]
   87 |     int ans[65];
      |         ^~~

• Subtask 1: 0 / 17
• Subtask 2: 0 / 17
• Subtask 3: 0 / 18
• Subtask 4: 0 / 29
• Subtask 5: 0 / 19