| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1063927 | thinknoexit | Alice, Bob, and Circuit (APIO23_abc) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "abc.h"
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
// 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
int lexico(char s[]) {
int len = strlen(s);
int res = 0;
for (int j = 0;j < len;j++) {
res = res * 26 + s[j] - 'a';
}
for (int j = 0;j < 4 - len;j++) res *= 26;
return res;
}
// Alice (returns la)
int alice(
const int n,
const char names[][5],
const unsigned short numbers[],
bool outputs_alice[]
) {
vector<int> rnk(n);
vector<pair<int, int>> v;
for (int i = 0;i < n;i++) {
rnk[i] = lexico(names[i]);
v.push_back({ rnk[i], i });
}
sort(v.begin(), v.end());
vector<int> ord(n);
for (int i = 0;i < n;i++) {
ord[v[i].second] = i;
}
for (int i = 0;i < n;i++) {
for (int j = 0;j < 16;j++)
outputs_alice[16 * i + j] = (numbers[v[i].second] >> j) & 1;
}
int sz = 16 * n;
// sort by lexicography order use 5 bit
for (int i = 0;i < n;i++) {
for (int j = 0;j < n;j++) {
outputs_alice[sz++] = (j == ord[i]);
}
}
return sz; // 16 * n + n * n
}
// Bob (returns lb)
int bob(
const int m,
const char senders[][5],
const char recipients[][5],
bool outputs_bob[]
) {
vector<int> r1(m), r2(m);
set<int> s;
for (int i = 0;i < m;i++) {
r1[i] = lexico(senders[i]);
r2[i] = lexico(recipients[i]);
s.insert(r1[i]), s.insert(r2[i]);
}
int n = s.size();
vector<int> v;
vector<vector<bool>> adj(n, vector<bool>(n, false));
for (auto& x : s) v.push_back(x);
for (int i = 0;i < m;i++) {
int o1 = lower_bound(v.begin(), v.end(), r1[i]) - v.begin();
int o2 = lower_bound(v.begin(), v.end(), r2[i]) - v.begin();
adj[o2][o1] = 1;
}
for (int i = 0;i < n;i++) {
for (int j = 0;j < n;j++) {
outputs_bob[i * n + j] = adj[i][j];
}
}
return n * n; // n * n
}
// Circuit (return l >= la + lb)
int circuit(
const int la,
const int lb,
int operations[],
int operands[][2],
int outputs_circuit[][16]
) {
int sz = la + lb;
int n = (la - lb) / 16;
// make one and zero
operations[sz] = OP_ZERO, operands[sz][0] = operands[sz][1] = 0;
const int ZERO = sz++;
// initialize
auto op = [&](int o, int a, int b) {
operations[sz] = o;
operands[sz][0] = a;
operands[sz][1] = b;
return sz++;};
auto sum = [&](int a[], int b[], int c[]) {
vector<int> s(16, 0);
c[0] = op(OP_XOR, a[0], b[0]);
s[0] = op(OP_AND, a[0], b[0]);
for (int j = 1;j < 16;j++) {
c[j] = op(OP_XOR, s[j - 1], op(OP_XOR, a[j], b[j]));
s[j] = op(OP_AND, op(OP_OR, s[j - 1], a[j]),
op(OP_AND, op(OP_OR, s[j - 1], b[j]), op(OP_OR, a[j], b[j])));
}};
//----------------
int tp[16] = {}, t[16] = {}, cand[n][16] = {};
for (int i = 0;i < n;i++) for (int j = 0;j < 16;j++) {
outputs_circuit[i][j] = cand[i][j] = ZERO;
}
for (int i = 0;i < n;i++) {
for (int j = 0;j < n;j++) {
for (int k = 0;k < 16;k++) t[k] = op(OP_AND, la + i * n + j, j * 16 + k);
sum(cand[i], t, tp);
for (int k = 0;k < 16;k++) cand[i][k] = tp[k];
}
}
for (int i = 0;i < n;i++) {
for (int j = 0;j < n;j++) {
for (int k = 0;k < 16;k++) t[k] = op(OP_AND, 16 * n + i * n + j, cand[j][k]);
sum(outputs_circuit[i], t, tp);
for (int k = 0;k < 16;k++) outputs_circuit[i][k] = tp[k];
}
}
return sz;
}