#pragma GCC optimize("O1,O2,O3,Ofast,unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
const char nl = '\n';
void fastIO() {
ios::sync_with_stdio(false);
cin.tie(0);
}
int N, T;
double P;
char query(vi v) {
string s(N, '0');
for (int x : v)
s[x] = '1';
cout<<"Q "<<s<<endl;
cout.flush();
char c;
cin>>c;
return c;
}
char query(int l, int r) {
string s(N, '0');
for (int i = l; i <= r; i++) {
s[i] = '1';
}
cout<<"Q "<<s<<endl;
cout.flush();
char c;
cin>>c;
return c;
}
string ans(N, '0');
void solve(int left, int right) { // know there's positive on range
if (left == right) {
ans[left] = '1';
return;
}
int mid1 = (2 * left + right) / 3;
int mid2 = (left + 2 * right) / 3;
if (query(left, mid1) == 'P') {
solve(left, mid1);
if (mid1 + 1 <= mid2 && query(mid1 + 1, mid2) == 'P')
solve(mid1 + 1, mid2);
if (mid2 + 1 <= right && query(mid2 + 1, right) == 'P')
solve(mid2 + 1, right);
}
else { // surely has in mid1 to right
solve(mid1 + 1, mid2);
if (mid2 + 1 <= right && query(mid2 + 1, right) == 'P')
solve(mid2 + 1, right);
}
}
int main() {
fastIO();
cin>>N>>P>>T;
int len = max(5, min((int)(0.8 / P), N));
for (int i = 0; i < T; i++) {
ans = string(N, '0');
for (int j = 0; j < N; j += len) {
int lp = j;
int rp = min(j + len - 1, N - 1);
if (query(lp, rp) == 'P')
solve(lp, rp);
}
// cout<<"ANSWER: ";
cout<<"A "<<ans<<endl;
cout.flush();
char c;
cin>>c;
if (c == 'W')
break;
}
}
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Execution timed out |
5430 ms |
46672 KB |
Time limit exceeded (wall clock) |
2 |
Halted |
0 ms |
0 KB |
- |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Execution timed out |
3625 ms |
26872 KB |
Time limit exceeded (wall clock) |
2 |
Halted |
0 ms |
0 KB |
- |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Execution timed out |
3442 ms |
27020 KB |
Time limit exceeded (wall clock) |
2 |
Halted |
0 ms |
0 KB |
- |