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;
int a[110], b[110];
int n;
int minValue(int N, int W) {
n=N;
// TODO: Implement Subtask 1 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
a[0]=1;
playRound(a, b);
for (int i=1; i<N; ++i) if (b[i]==0) return i;
return 0;
}
int maxValue(int N, int W) {
n=N;
// TODO: Implement Subtask 2 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
for (int i=0; i<N; ++i){
a[i]=1;
}
playRound(a, b);
// for (int i=0; i<N; ++i) if (b[i]) cout << i+1 << ' ';
// cout << '\n';
for (int i=0; i<N; ++i){
a[i]=(b[i]>a[i] && a[i])?2:0;
}
playRound(a, b);
// for (int i=0; i<N; ++i) if (b[i]) cout << i+1 << ' ';
// cout << '\n';
for (int i=0; i<N; ++i){
a[i]=(b[i]>a[i] && a[i])?4:0;
}
playRound(a, b);
// for (int i=0; i<N; ++i) if (b[i]) cout << i+1 << ' ';
// cout << '\n';
for (int i=0; i<N; ++i){
a[i]=(b[i]>a[i] && a[i])?11:0;
}
playRound(a, b);
for (int i=0; i<N; ++i) if (b[i]>a[i] && a[i]) return i;
// for (int i=0; i<N; ++i) if (b[i]) cout << i+1 << ' ';
// cout << '\n';
return 0;
}
int ans[110], ans2[110];
int greaterValue(int N, int W) {
n=N;
// TODO: Implement Subtask 3 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
for (int i=0; i<N; ++i){
a[i]=1;
}
playRound(a, b);
if ((b[0]>a[0])!=(b[1]>a[1])){
if (b[0]>a[0]) return 0;
return 1;
}
if (b[0]>a[0] && b[1]>a[1]){
for (int i=0; i<N; ++i) a[i]=(i==0 || i==1)?10:0;
playRound(a, b);
if (b[0]>a[0]) return 0;
return 1;
}
// bitset<1225> bs[11];
// for (int i=0; i<=10; ++i) bs[i].reset();
// for (int v=1; v<=10; ++v){
// int cur=0;
// for (int x=0; x<N; ++x) for (int y=x+1; y<N; ++y) if (x<50 && y<50){
// for (int i=0; i<N; ++i) a[i]=(i==x || i==y)?v:0;
// playRound(a, b);
// bs[v].set(cur++, b[x]!=b[y]);
// }
// }
// for (int i=0; i<1024; ++i){
// bitset<1225> tot;
// for (int j=0; j<10; ++j) if (i>>j&1) tot|=bs[j+1];
// if (tot.all()){
// for (int j=0; j<10; ++j) if (i>>j&1) cout << j+1 << ' ';
// cout << '\n';
// }
// }
for (int v:{1, 3, 6}){
for (int i=0; i<N; ++i) a[i]=(i==0 || i==1)?v:0;
playRound(a, b);
if ((b[0]>a[0])!=(b[1]>a[1])){
if (b[0]>a[0]) return 0;
return 1;
}
}
return 0;
}
bool compare(int x, int y){
if (ans[x]!=ans[y]){
if (ans[x]) return 0;
return 1;
}
if (ans[x] && ans[y]){
for (int i=0; i<n; ++i) a[i]=(i==x || i==y)?10:0;
playRound(a, b);
if (b[x]>a[x]) return 0;
return 1;
}
for (int v:{6, 3, 1}){
for (int i=0; i<n; ++i) a[i]=(i==x || i==y)?v:0;
playRound(a, b);
if ((b[x]>a[x])!=(b[y]>a[y])){
if (b[x]>a[x]) return 0;
return 1;
}
}
return 0;
}
void allValues(int N, int W, int *P) {
n=N;
if (W == 2*N) {
// TODO: Implement Subtask 4 solution here.
// You may leave this block unmodified if you are not attempting this
// subtask.
auto comp=[&](int x, int y) -> bool {
for (int i=0; i<N; ++i) a[i]=(i==x || i==y)?100:0;
playRound(a, b);
return (b[y]>a[y]);
};
auto merge_sort=[&](auto self, vector<int> &v) -> void {
int sz=(int)v.size();
if (sz<=1) return;
vector<int> vl(v.begin(), v.begin()+sz/2);
vector<int> vr(v.begin()+sz/2, v.end());
self(self, vl);
self(self, vr);
merge(vl.begin(), vl.end(), vr.begin(), vr.end(), v.begin(), comp);
};
vector<int> v(N);
iota(v.begin(), v.end(), 0);
merge_sort(merge_sort, v);
for (int i=0; i<N; ++i) P[v[i]]=i+1;
} else {
// TODO: Implement Subtask 5 solution here.
// You may leave this block unmodified if you are not attempting this
// subtask.
int lval=1;
auto dnc=[&](auto self, vector<int> &v) -> void {
int sz=(int)v.size();
if (sz<=1){
++lval;
return;
}
vector<int> vl;
vector<int> vr;
for (int i=0; i<n; ++i) a[i]=0;
for (int i:v) a[i]=min((int)sqrt(2*lval), 100/sz);
playRound(a, b);
for (int i:v) if (b[i]>a[i]) vr.push_back(i); else vl.push_back(i);
self(self, vl);
self(self, vr);
v.clear();
v.insert(v.end(), vl.begin(), vl.end());
v.insert(v.end(), vr.begin(), vr.end());
};
vector<int> v(n); iota(v.begin(), v.end(), 0);
dnc(dnc, v);
for (int i=0; i<N; ++i) P[v[i]]=i+1;
}
}
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