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 "koala.h"
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define double long double
using pii = pair<int, int>;
template<typename T>
using prior = std::priority_queue<T, vector<T>, greater<T>>;
template<typename T>
using Prior = std::priority_queue<T>;
#define X first
#define Y second
#define ALL(x) (x).begin(), (x).end()
#define eb emplace_back
#define pb push_back
#define fastIO() ios_base::sync_with_stdio(false), cin.tie(0)
int minValue(int N, int W) {
// TODO: Implement Subtask 1 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
int play[N], ret[N];
memset(play, 0x00, sizeof(play));
play[0] = 1;
playRound(play, ret);
for (int i = 0; i < N; ++i) {
if (ret[i] <= play[i]) return i;
}
return 0;
}
int maxValue(int N, int W) {
// TODO: Implement Subtask 2 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
int play[N], ret[N];
memset(play, 0x00, sizeof(play));
set<int> test;
for (int i = 0; i < N; ++i) test.insert(i);
while (test.size() != 1) {
int cookie = W / test.size();
memset(play, 0x00, sizeof(play));
for (auto x : test) play[x] = cookie;
playRound(play, ret);
vector<int> tmp;
for (auto x : test) {
if (ret[x] <= play[x]) tmp.eb(x);
}
for (auto x : tmp) test.erase(x);
}
return *test.begin();
}
set<int> known;
int maxValue2(int N, int W) {
int play[N], ret[N];
memset(play, 0x00, sizeof(play));
set<int> test;
for (int i = 0; i < N; ++i) test.insert(i);
for (auto x : known) test.erase(x);
while (test.size() != 1) {
int cookie = W / test.size();
memset(play, 0x00, sizeof(play));
for (auto x : test) play[x] = cookie;
playRound(play, ret);
vector<int> tmp;
for (auto x : test) {
if (ret[x] <= play[x]) tmp.eb(x);
}
for (auto x : tmp) test.erase(x);
}
return *test.begin();
}
int greaterValue(int N, int W) {
// TODO: Implement Subtask 3 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
int play[N], ret[N];
memset(play, 0x00, sizeof(play));
play[0] = play[1] = 3;
playRound(play, ret);
if (ret[0] != ret[1]) return ret[1] > ret[0];
if (ret[0] == 0) {
play[0] = play[1] = 1;
playRound(play, ret);
if (ret[0] != ret[1]) return ret[1] > ret[0];
play[0] = play[1] = 2;
playRound(play, ret);
if (ret[0] != ret[1]) return ret[1] > ret[0];
}
else {
play[0] = play[1] = 5;
playRound(play, ret);
if (ret[0] != ret[1]) return ret[1] > ret[0];
play[0] = play[1] = 8;
playRound(play, ret);
if (ret[0] != ret[1]) return ret[1] > ret[0];
}
return 0;
}
void allValues(int N, int W, int *P) {
if (W == 2*N) {
// TODO: Implement Subtask 4 solution here.
// You may leave this block unmodified if you are not attempting this
// subtask.
for (int i = 0; i < N; ++i) {
int mV = maxValue2(N, W);
*(P + mV) = N - i;
known.insert(mV);
}
} else {
// TODO: Implement Subtask 5 solution here.
// You may leave this block unmodified if you are not attempting this
// subtask.
for (int i = 0; i < N; ++i) {
int mV = maxValue2(N, W);
*(P + mV) = N - i;
known.insert(mV);
}
}
}
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