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 "combo.h"
#include <vector>
using namespace std;
string opt[] = {"A", "B", "X", "Y"};
vector<string> oth;
string find_first() {
string p = "";
int tre = 0;
bool ok = 0;
for (string c : opt) {
if (p.length() > 0 && c == string(1, p[0])) continue;
if (ok || press(p + c) == (int)p.length() + 1) {
p += c;
break;
} else tre++;
if (p.length() > 0 && tre == 2) ok = 1;
if (p.length() == 0 && tre == 3) ok = 1;
}
for (string c : opt) {
if (string(1, p[0]) != c) oth.push_back(c);
}
return p;
}
string find_two(string sd) {
vector<string> cmb;
for (string c : oth) {
cmb.push_back(oth[0] + c);
}
cmb.push_back(oth[1] + oth[0]);
string q;
for (auto x : cmb) q += sd + x;
int out = press(q) - sd.length();
if (out == 2) {
q.clear();
q = sd + cmb[0] + sd + cmb[1];
out = press(q) - sd.length();
if (out == 2) {
out = press(sd + cmb[0]) - sd.length();
if (out == 2) {
return cmb[0];
} else return cmb[1];
} else if (out == 1) return cmb[2];
else return cmb[3];
} else if (out == 1) {
out = press(sd + oth[1] + oth[1]) - sd.length();
if (out == 2) return oth[1] + oth[1];
else return oth[1] + oth[2];
} else {
return oth[2];
vector<string> opt;
for (int i = 0; i < 3; ++i) opt.push_back(oth[2] + oth[i]);
for (int i = 0; i < 2; ++i) {
if (press(sd + opt[i]) == (int)sd.length() + 2) return opt[i];
}
return opt[2];
}
}
string find_one(string p) {
int tre = 0;
bool ok = 0;
for (string c : opt) {
if (p.length() > 0 && c == string(1, p[0])) continue;
if (ok || press(p + c) == (int)p.length() + 1) {
p += c;
break;
} else tre++;
if (p.length() > 0 && tre == 2) ok = 1;
if (p.length() == 0 && tre == 3) ok = 1;
}
return p;
}
string guess_sequence(int N) {
oth.clear();
string p = find_first();
while ((int)p.length() + 2 <= N) {
p += find_two(p);
}
if ((int)p.length() != N) {
p = find_one(p);
}
return p;
}
/*
S'AA S'AB S'AX S'BA
2 -> Jokin näistä
-> 2/1 kyselyä vielä = 3/2
1 -> Alkaa B, mutta ei ole BA
-> 1 kysely vielä = 2
0 -> Ei ala B, eikä ole A_
-> 0 enää kyselyä ja saadaan 1 merkki
S'AA S'AB S'BA S'BB
2 -> Jokin näistä
-> Kysy S'
*/
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