| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 982224 | 8pete8 | 앨리스, 밥, 서킷 (APIO23_abc) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "abc.h"
#include<bits/stdc++.h>
// you may find the definitions useful
const int OP_ZERO = 0; // f(OP_ZERO, x0, x1) = 0
const int OP_NOR = 1; // f(OP_NOR, x0, x1) = !(x0 || x1)
const int OP_GREATER = 2; // f(OP_GREATER, x0, x1) = (x0 > x1)
const int OP_NOT_X1 = 3; // f(OP_NOT_X1, x0, x1) = !x1
const int OP_LESS = 4; // f(OP_LESS, x0, x1) = (x0 < x1)
const int OP_NOT_X0 = 5; // f(OP_NOT_X0, x0, x1) = !x0
const int OP_XOR = 6; // f(OP_XOR, x0, x1) = (x0 ^ x1)
const int OP_NAND = 7; // f(OP_NAND, x0, x1) = !(x0 && x1)
const int OP_AND = 8; // f(OP_AND, x0, x1) = (x0 && x1)
const int OP_EQUAL = 9; // f(OP_EQUAL, x0, x1) = (x0 == x1)
const int OP_X0 = 10; // f(OP_X0, x0, x1) = x0
const int OP_GEQ = 11; // f(OP_GEQ, x0, x1) = (x0 >= x1)
const int OP_X1 = 12; // f(OP_X1, x0, x1) = x1
const int OP_LEQ = 13; // f(OP_LEQ, x0, x1) = (x0 <= x1)
const int OP_OR = 14; // f(OP_OR, x0, x1) = (x0 || x1)
const int OP_ONE = 15; // f(OP_ONE, x0, x1) = 1
// Alice
int // returns la
alice(
/* in */ const int n,
/* in */ const char names[][5],
/* in */ const unsigned short numbers[],
/* out */ bool outputs_alice[]
){
//nume of char x4
//we will get a 10^8 value? then hash the bit
//24 bit
//each person =24+16=40;
vector<pair<string,int>>v;
return 0;
int mx=40*n;
for(int i=0;i<mx;i++)outputs_alice[i]=0;
int st=0;
for(int i=0;i<n;i++){
int sum=0;
int base=1;
for(int j=0;j<4;j++){
char a=names[i][j];
int val=0;
if(a>='a'&&a<='z')val='a'-a;
else val=26;
sum+=(base*(val%10));
base*=10;
val/=10;
sum+=(base*(val%10));
base*=10;
}
for(int j=0;j<24;j++)if(sum&(1<<j))outputs_alice[st+j]=1;
st+=24;
for(int j=0;j<16;j++)if(numbers[i]&(1<<j))outputs_alice[st+j]=1;
st+=16;
}
return mx;
//block is 0->124
}
// Bob
int // returns lb
bob(
/* in */ const int m,
/* in */ const char senders[][5],
/* in */ const char recipients[][5],
/* out */ bool outputs_bob[]
) {
int st=0;
//use name hashx2x900
//24x2x900
for(int i=0;i<(24*2*m);i++)outputs_bob[i]=0;
for(int i=0;i<m;i++){
int sum=0;
int base=1;
for(int j=0;j<4;j++){
char a=senders[i][j];
int val=0;
if(a>='a'&&a<='z')val='a'-a;
else val=26;
sum+=(base*(val%10));
base*=10;
val/=10;
sum+=(base*(val%10));
base*=10;
}
for(int j=0;j<24;j++)if(sum&(1<<j))outputs_bob[st+j]=1;
st+=24;
sum=0;
base=1;
for(int j=0;j<4;j++){
char a=recipients[i][j];
int val=0;
if(a>='a'&&a<='z')val='a'-a;
else val=26;
sum+=(base*(val%10));
base*=10;
val/=10;
sum+=(base*(val%10));
base*=10;
}
for(int j=0;j<24;j++)if(sum&(1<<j))outputs_bob[st+j]=1;
st+=24;
}
return (24*2*m);
}
// Circuit
int // returns l
circuit(
/* in */ const int la,
/* in */ const int lb,
/* out */ int operations[],
/* out */ int operands[][2],
/* out */ int outputs_circuit[][16]
) {
//test
int id=la+lb;
int n=la/(40),m=lb/24;
for(int j=0;j<n;j++)for(int i=0;i<16;i++)outputs_circuit[j][i]=-1;
auto add=[&](int a,int b,int x){
operations[id]=x;
operands[id][0]=a,operands[id][1]=b;
id++;
return id-1;
};
auto actualadd=[&](int pos,int x,int who){//adding to "x" to bit "pos" of "who"
int last=x;
for(int k=pos;k<16;k++){
if(outputs_circuit[who][k]==-1){
outputs_circuit[who][k]=last;
break;
}
int nx=add(last,outputs_circuit[who][k],OP_AND);
outputs_circuit[who][k]=add(outputs_circuit[who][k],last,OP_XOR);
last=nx;
}
};
for(int i=0;i<m;i++){//900
int st=la;
//m*(n*hashname*2)*(n*hashname*2+32+16x16)<2e7;
for(int j=0;j<n;j++){//30
int id=(40*j);
int x=add(id,st,OP_EQUAL);
for(int g=1;g<24;g++){//24
x=add(x,add(id+g,st+g,OP_EQUAL),OP_AND);//2
}
for(int k=j;k<n;k++){
int id2=(40*k);
int y=add(id2,st+24,OP_EQUAL);
for(int g=1;g<24;g++){//24
y=add(y,add(id2+g,st+24+g,OP_EQUAL),OP_AND);//2
}
for(int g=0;g<16;g++)actualadd(g,add(id+24+g,add(x,y,OP_AND),OP_AND),k);
}
}
st+=24;
/*
wtf did i wrote TT this not gonna pass
*/
}
return id;
/*
when we are multiplying
if we look at it bit by bit
a,b =shift a->a+b
if its >16 then it will have to be shifted down
but what if that bit is arldy on
then we can find the max consecutive 1s bit and turn the pos+1 to 1
then make everything there 0
if we know the sender and reciever
then we add reciever bit
*/
}