# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
1052570 | Piokemon | 앵무새 (IOI11_parrots) | C++17 | 4 ms | 1644 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "encoder.h"
#include "encoderlib.h"
using namespace std;
#include <bits/stdc++.h>
typedef __int128 ll;
void encode(int n, int m[]){
ll ilosc[42][35];
// poz ter
for (int x=0;x<=40;x++){
for (int y=0;y<=33;y++)ilosc[x][y]=0;
}
ilosc[1][1]=1;
for (int x=1;x<=40;x++){
for (int y=1;y<=33;y++){
ilosc[x][y+1]+=ilosc[x][y];
ilosc[x+1][y]+=ilosc[x][y];
}
}
vector<int> dupa;
for (int blok=0;blok*8<n;blok++){
int dlug=min(8,n-blok*8);
ll wart=0;
for (int x=0;x<dlug;x++){
//cerr << (int)wart << ' ' << (int)wart*256 << ' ' << m[blok*8+x] << '\n';
if (blok*8+x>=n)wart=wart*256;
else wart=(wart*256)+(ll)m[blok*8+x];
}
int nr=0; //rang od blok*32 do blok*32+31;
for (int x=0;x<dlug*5;x++){
while (ilosc[dlug*5-x][32-nr]<=wart){
wart-=ilosc[dlug*5-x][32-nr];
nr++;
}
dupa.push_back(nr+blok*32);
}
}
//for (int x:dupa) cerr << "send " << x << '\n';
for (int x:dupa)send(x);
}
#include "decoder.h"
#include "decoderlib.h"
using namespace std;
#include <bits/stdc++.h>
typedef __int128 ll;
void decode(int n, int l, int x[])
{
ll ilosc[42][35];
// poz ter
for (int z=0;z<=40;z++){
for (int y=0;y<=33;y++)ilosc[z][y]=0;
}
ilosc[1][1]=1;
for (int z=1;z<=40;z++){
for (int y=1;y<=32;y++){
ilosc[z][y+1]+=ilosc[z][y];
ilosc[z+1][y]+=ilosc[z][y];
}
}
sort(x,x+l);
vector<int> dupa2;
for (int blok=0;blok*8<n;blok++){
int dlug=min(8,n-blok*8);
// bieremy papugi od blok*40 do blok*40+39
ll wart=0;
int nr=0;
for (int y=0;y<5*dlug;y++){
x[blok*40+y]-=blok*32;
while(x[blok*40+y]>nr){
wart+=ilosc[5*dlug-y][32-nr];
nr++;
}
}
/*cerr << "odeb ";
vector<int> tempo;
ll kop=wart;
while(kop){
tempo.push_back(kop%(ll)10);
kop/=(ll)10;
}
for (int y=tempo.size()-1;y>=0;y--) cerr << tempo[y];
cout << '\n';*/
vector<int> temp;
for (int y=0;y<dlug;y++){
temp.push_back(wart%(ll)256);
wart/=(ll)256;
}
for (int y=dlug-1;y>=0;y--)dupa2.push_back(temp[y]);
}
while(dupa2.size()>n)dupa2.pop_back();
//for (int y:dupa2) cerr << "out " << y << '\n';
for (int y:dupa2) output(y);
}
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